Prep - 0.1 First Order Differential Equations 1. Solve the...

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Unformatted text preview: 0.1 First Order Differential Equations 1. Solve the following separable differential equations. −2x (a) y = e y . We start by separating the equation: dy e−2x = dx y y dy = e−2x dx Next, we integrate both sides: y dy = e−2x dx y2 e−2x =− +C 2 2 Lastly, solving for y : y2 e−2x =− +C 2 2 y 2 = −e−2x + 2C y = ± −e−2x + 2C = ± −e−2x + D D = 2C. (b) 3y 5 y − x2 (x3 + 5)10 = 0 We start by separating the equation: dy = x2 (x3 + 5)10 dx 3y 5 dy = x2 (x3 + 5)10 dx 3y 5 Next, we integrate both sides: 3y 5 dy = x2 (x3 + 5)10 dx Let u = x3 + 5, so du = 3x2 dx: y6 = 2 1 10 u du 3 1 11 u +C 33 13 = (x + 5)11 + C 33 = Lastly, solving for y : y6 13 = (x + 5)11 + C 2 33 23 23 y6 = (x + 5)11 + 2C = (x + 5)11 + D 33 33 1 D = 2C. x − (c) y = ey+22 . We start by separating the equation: ex − 2 dy = dx y+2 (y + 2) dy = ex − 2 dx Next, we integrate both sides: (y + 2) dy = ex − 2 dx y2 + 2y = ex − 2x + C 2 Lastly, we solve for y using the quadratic formula: y2 + 2y − (ex − 2x + C ) = 0 2 y = −2 ± 4 − 2(ex − 2x + C ) √ = −2 ± 4 − 2ex − 4x + D D = 2C. (d) y = − x1+2y ) . (1+y We start by separating the equation: dy 1 + 2y =− dx x(1 + y ) 1+y 1 dy = − dx 1 + 2y x Next, we integrate both sides: 1 1+y dy = − dx 1 + 2y x 1 1 1 2 (1 + 2y ) + 2 dy = − dx 1 + 2y x 111 1 + dy = − dx 2 2 1 + 2y x 1 1 y + ln |1 + 2y | = − ln x + C 2 4 Here, we cannot solve for y , so we leave the solution in implicit form. 2 (e) y + x2 y = x2 y 2 . We start by separating the equation: dy = x2 y 2 − x2 y = x2 ( y 2 − y ) dx dy = x2 dx y2 − y Next, we integrate both sides: dy = y2 − y x2 dx. We solve the left hand side by partial fractions: y2 1 1 1 =− + , −y y y−1 so − ln |y | + ln |y − 1| = y−1 |= y 1 ln |1 − | = y ln | x3 +C 3 x3 +C 3 x3 +C 3 Lastly, we solve for y : x3 x3 1 |1 − | = e 3 +C = De 3 y x3 1 1 − = ±De 3 y 1 y= x3 1 ± De 3 3 D = eC 2. Solve the following homogeneous equations. (a) y = y 5 −x5 y4 x . Start by checking that the equation is homogeneous: let f (x, y ) = y 5 −x 5 y 4 x . Then, f (xt, yt) = Let y = vx, so (yt)5 − (xt)5 t5 (y 5 − x5 ) y 5 − x5 = = = f (x, y ). (yt)4 (xt) t5 (y 4 x) y4 x dy dx v+x dv = v + x dx . Substitute these into the equation: v5 − 1 dv ((vx)5 − x5 x5 (v 5 − 1) = = = dx (vx)4 x x5 v 4 v4 5 dv v −1 1 x = − v = − 4. 4 dx v v The new equation is separable: v 4 dv = − v 4 dv = − dx x dx x v5 = − ln |x| + C 5 Let C = − ln |k |, so v5 = − ln |x| − ln |k | = − ln |kx| 5 v 5 = −5 ln |kx| = ln |kx|−5 √ v = 5ln |kx|−5 Lastly, writing the expression in terms of y gives: √ y = vx = x 5ln |kx|−5 . 4 (b) y = −2y +3x . y Start by checking that the equation is homogeneous: let f (x, y ) = −2y +3x . Then, y f (xt, yt) = Let y = vx, so dy dx −2(yt) + 3(xt) −2y + 3x = = f (x, y ). (yt) y dv = v + x dx . Substitute these into the equation: dv −2(vx) + 3x −2v + 3 = = dx vx v dv −2v + 3 −v 2 − 2v + 3 x = −v = dx v v dx v dv = − v 2 + 2v − 3 x dx v dv = − 2 + 2v − 3 v x v+x The right hand side can be done by partial fractions: v2 v 31 11 = + , + 2v − 3 4v+3 4v−1 so 11 dx 31 + dv = − 4v+3 4v−1 x 3 1 ln |v + 3| + ln |v − 1| = − ln |x| + C 4 4 Let C = − ln |k |, so 3 1 ln |v + 3| + ln |v − 1| = − ln |x| − ln |k | = − ln |kx| 4 4 3 ln |v + 3| + ln |v − 1| = −4 ln |kx| ln |v + 3|3 + ln |v − 1| = ln |kx|−4 ln |v + 3|3 |v − 1| = ln |kx|−4 |v + 3|3 |v − 1| = |kx|−4 y y | + 3|3 | − 1| = |kx|−4 x x This expression cannot be simplified further, so we leave it as is. 5 (c) y = x2 −y 2 xy . Start by checking that the equation is homogeneous: let f (x, y ) = x 2 −y 2 xy . Then, f (xt, yt) = Let y = vx, so dy dx x2 − y 2 (xt)2 − (yt)2 = = f (x, y ). (xt)(yt) xy dv = v + x dx . Substitute these into the equation: dv x2 − (vx)2 1 − v2 = = dx x(vx) v dv 1 − v2 1 − 2v 2 x = −v = dx v v dx v dv = 1 − 2v 2 x v dx dv = 1 − 2v 2 x v+x Let u = 1 − 2v 2 , so du = −4v dv and − du = v dv : 4 − 1 4 du = u dx x 1 − ln |u| = ln |x| + C. 4 Let C = ln |k |, so 1 − ln |u| = ln |kx| 4 1 − ln |1 − 2v 2 | = ln |kx| 4 1 ln |1 − 2v 2 |− 4 = ln |kx| 1 |1 − 2v 2 |− 4 = |kx| |1 − 2v 2 | = |kx|−4 1 − 2v 2 = ±|kx|−4 1 v2 = 1 ± |kx|−4 2 1 v=± (1 ± |kx|−4 ) 2 1 y = ±x (1 ± |kx|−4 ) 2 6 (d) y = x3 +y 3 xy 2 . Start by checking that the equation is homogeneous: let f (x, y ) = x3 +y 3 xy 2 . Then, f (xt, yt) = Let y = vx, so dy dx x3 + y 3 (xt)3 + (yt)3 = = f (x, y ). 2 (xt)(yt) xy 2 dv = v + x dx . Substitute these into the equation: 1 + v3 dv x3 + (vx)3 = = dx x(vx)2 v2 dv 1 + v3 1 x = −v = 2 2 dx v v dx 2 v dv = x dx 2 v dv = x 3 v = ln |x| + C 3 v+x Let C = ln |k |, so v3 = ln |kx| 3 3 v = 3 ln |kx| = ln |kx|3 v= 3 ln |kx|3 y = x 3 ln |kx|3 . 7 (e) y 2 + x2 y = xyy . Start by writing this in the form y = f (x, y ): x2 y − xyy = −y 2 (x2 − xy )y = −y 2 y =− y2 x2 − xy Next, check that the equation is homogeneous: f (xt, yt) = − Let y = vx, so dy dx (xt)2 (yt)2 y2 =− 2 = f (x, y ). − (xt)(yt) x − xy dv = v + x dx . Substitute these into the equation: dv (vx)2 v2 =− 2 = dx x − x(vx) 1−v dv v2 v x = −v = dx 1−v v−1 v−1 dx dv = v x v−1 dx dv = v x 1 dx 1 − dv = v x v − ln |v | = ln |x| + C v+x Let C = ln |k |, so v − ln |v | = − ln |kx|. This expression cannot be simplfied further, so we simply substitute y v = x: y y − ln | | = ln |kx|. x x 8 3. Find the function y = y (x) such that dy dx = y 2 (1 + 2x) and y (0) = −1. Separating variables and integrating yields: dy = 1 + 2x dx y2 dy = 1 + 2x dx y2 1 − = x + x2 + C y 1 y=− x + x2 + C Now, we have −1 = y (0) = − so C = 1 and y=− 1 , C 1 . x + x2 + 1 4. Solve the following first-order linear differential equations. (a) dy dx y + x − x5 = 0 This is first-order linear, with P (x) = solution y = e− = e− P (x) dx dx x = e− ln x e e dx x 1 x P (x) dx 1 x6 dx + C x 1 17 x +C = x7 C 1 = x6 + 7 x 9 Q(x) dx + C x5 dx + C eln x x5 dx + C = and Q(x) = x5 , so it has (b) dy dx + y = sin x Here, P (x) = 1 and Q(x) = sin x, so y = e− dx = e−x = e−x dx e sin x dx + C ex sin x dx + C ex sin x − ex cos x +C 2 (see section 1.1.5 for this integral.) = (c) dy dx − y x+1 sin x − cos x + Ce−x 2 = x2 1 Here, P (x) = − 1+x and Q(x) = x2 , so y=e 1 1+x dx = eln (1+x) = (1 + x) = (1 + x) = (1 + x) = (1 + x) = (1 + x) e− 1 1+x dx 2 x dx + C e− ln (1+x) x2 dx + C x2 dx + C 1+x (x2 − 1) + 1 dx + C 1+x (1 + x)(1 − x) + 1 dx + C 1+x 1 1−x+ dx + C 1+x x2 x− + ln |1 + x| + C 2 10 (d) xy + y = 3x cos 2x. Rearranging, we get y+ 1 y = 3 cos 2x, x 1 so P (x) = − x and Q(x) = 3 cos 2x. Thus, y = e− 1 x dx 3e = e− ln x = We integrate 1 x 1 x dx cos 2x dx + C 3eln x cos 2x dx + C x cos 2x dx + C x cos 2x dx by parts: u = 3x dv = cos 2x dx 1 du = 3 dx v = sin 2x 2 Thus, x cos 2x dx = 3 3 x sin 2x − 2 2 sin 2x dx = 3 3 x sin 2x + cos 2x, 2 4 so, 13 3 x sin 2x + cos 2x + C x2 4 3 3 C = sin 2x + cos 2x + . 2 4x x y= 11 5. Solve the following linear differential equations. Use the initial conditions given to find all unknown constants. (a) y − 3xy = 15x, y (0) = 5. Here, P (x) = −3x and Q(x) = 15x, so 15xe− 3x dx y=e 2 3 3 3x dx dx + C 2 15xe− 2 x dx + C . = e2x 3 Let u = − 2 x2 , so du = −3x dx and − du = x dx: 3 3 2 15 3 = −5eu 15xe− 2 x dx = − eu du 3 2 = −5e− 2 x Thus, 3 y = e2x 2 3 2 −5e− 2 x + C 3 = −5 + Ce 2 x 2 Lastly, we use the initial condition y (0) = 5 to find C : 5 = y (0) = −5 + C C = 10 3 2 Thus, y = −5 + 10e 2 x . 12 (b) y + 4xy = x, y (0) = 5 . 4 Here, P (x) = 4x and Q(x) = x, so y = e− xe− 4x dx = e−2x 2 4x dx dx + C 2 xe2x dx + C Let u = 2x2 , so du = 4x dx and du 4 = x dx: 1 eu du 4 1 = eu 4 12 = e2x 4 2 xe2x dx = Thus, y = e−2x = 2 1 2x2 e +C 4 2 1 + Ce−2x 4 Lastly, we use the initial condition y (0) = 5 4 5 1 = y (0) = + C 4 4 C=1 Thus, y = 1 4 2 + e−2x . 13 to find C : 1 (c) y + x y = ex , y (1) = 2. 1 Here, P (x) = x and Q(x) = ex , so 1 1 x y = e− ex eint x dx + C = e− ln x = Integrate 1 x ex eln x dx + C xex dx + C xex dx by parts: dv = ex dx u=x du = 1 v = ex Thus, xex dx = xex − ex dx = xex − ex , and 1 (xex − ex + C ) x ex = ex + +C x y= Lastly, we use the initial condition y (1) = 2 to find C : 2 = y (1) = e − e + C C = 2, so y = ex + ex x + 2. 14 6. Make a substitution to solve the following Bernoulli equations. If initial conditions are given, use them to solve for all unknown constants. (a) y + x2 y = x2 y 2 . 1 z Let z = y 1−2 = y −1 , so y = z and y = − z2 . Substitute these into the original equation and isolate for z : y + x2 y = x2 y 2 x2 z x2 + =2 z2 z z z − x2 z = −x2 − This is linear, with P (x) = −x2 and Q(x) = −x2 , so x2 dx z=e 1 = e3x 3 x2 e− − 1 x2 dx + C 3 x2 e− 3 x dx + C − 1 Let u = − 3 x3 , so du = −x2 dx: − and 1 3 x2 e− 3 x dx = 1 z = e3x 3 1 1 3 1 3 e− 3 x + C = 1 + Ce 3 x . Thus, y= 1 1 = 1 3. z 1 + Ce 3 x 15 3 eu du = eu = e− 3 x , √ (b) y + xy = x 3 y. 1 2 3 1 3 Let z = y 1− 3 = y 3 , so y = z 2 and y = 2 z 2 z . Substitute these into the original equation and isolate for z : 1 y + xy = xy 3 3 31 1 31 z 2 z + xz 2 = x(z 2 ) 3 = xz 2 2 2 2 z + xz = x 3 3 This is linear, with P (x) = 2 x and Q(x) = 2 x, so 3 3 2 3 x dx z = e− = e− Let u = x2 3, 2 2x xe 3 dx + C 3 2 x2 xe 3 dx + C 3 x2 3 so du = 2 x dx: 3 2 x2 xe 3 dx = 3 so z = e− x2 3 e x2 3 eu du = eu = e + C = 1 + Ce− and 2 y = z 3 = 1 + Ce− 16 x2 3 2 3 . x2 3 x2 3 , √ (c) y + 3x2 y = x2 y. 1 1 Let z = y 1− 2 = y 2 , so y = z 2 and y = 2zz . Substitute these into the original equation and isolate for z : 1 y + 3 x2 y = x2 y 2 2zz + 3x2 z 2 = x2 z x2 3 z + x2 z = 2 2 3 This is linear, with P (x) = 2 x2 and Q(x) = x2 , so 3 2 z = e− 2 x = e− Let u = x3 2, x2 3 x2 e 2 dx + C 2 dx x2 x3 e 2 dx + C 2 x3 2 so du = 3 x2 and 2 du 3 x2 x3 1 e 2 dx = 2 3 so z = e− x3 2 = x2 2 eu du = 1 x3 e 2 +C 3 and y = z2 = 17 dx: = 1u 1 x3 e = e2, 3 3 x3 1 + Ce− 2 , 3 x3 1 + Ce− 2 3 . 1 4 (d) y + x y = xy 4 , y (1) = 1. 1 3 4 1 4 Let z = y 1− 4 = y 4 , so y = z 3 and y = 3 z 3 z . Substitute these into the original equation and isolate for z : 1 4 y = xy 4 x 1 41 44 z 3 z + z 3 = xz 3 3 x 3 3 z + z= x x 4 y+ This is linear with P (x) = and Q(x) = 3 x, so 4 3 x 3 3 xe x dx + C 4 3 3 ln x xe dx + C 4 3 x z = e− = e−3 ln x 33 xx dx + C 4 34 x dx + C 4 = x −3 = x −3 = x −3 = 35 x +C 20 32 x + Cx−3 20 Thus, 4 3 32 x + Cx−3 20 4 y = z3 = . Lastly, we use the initial condition y (1) = 1 to find C : 4 1 = y (1) = 3 3 3 +C +C 1= 20 20 17 C= 20 Thus, y= 32 17 x+ 20 20x3 18 4 3 . 1 (e) y − x y = ex y 2 , y (1) = 1 . 2 z 1 Let z = y 1−2 = y −1 , so y = z and y = − z2 . Substitute these into the original equation and isolate for z : 1 y = ex y 2 x 11 z 1 − 2− = ex 2 z xz z 1 x z + z = −e x y− 1 x This is linear with P (x) = 1 x z = e− and Q(x) = −ex , so dx − = e− ln x − = 1 x 1 ex e x dx + C ex eln x dx + C xex dx + C − We do this by integration by parts, with u=x du = dx dv = ex dx v = ex , so xex dx = xex − and ex dx = xex − ex , z= 1 (−xex + ex + C ) . x y= 1 1 = . z −xex + ex + C Thus, Lastly, we use the initial condition y (1) = 1 2 to find C : 1 1 1 = y (1) = = 2 −e + e + C C C = 2, so y = 1 −xex +ex +2 . 19 7. Solve the following exact differential equations. (a) (y 2 cos x) dx + (y 2 + 2y sin x) dy = 0. Here M (x, y ) = y 2 cos x and N (x, y ) = y 2 + 2y sin x, and ∂M ∂N = 2y cos x = , ∂y ∂x so the equation is exact. Thus, g (x, y ) = M (x, y ) dx + h(y ) = y 2 cos x dx + h(y ) = y 2 sin x + h(y ) Since ∂g ∂y = N (x, y ), we must have y 2 + 2y sin x = Thus, h(y ) = y3 3 ∂ y 2 sin x + h(y ) = 2y sin x + h (y ) ∂y h (y ) = y 2 + C , so g (x, y ) = y 2 sin x + y 2 sin x + y 2 sin x + y3 +C =0 3 y3 =k 3 20 y3 +C 3 k = −C (b) (3x2 − 2xy + 2) dx + (6y 2 − x2 + 3) = 0 Here M (x, y ) = 3x2 − 2xy + 2 and N (x, y ) = 6y 2 − x2 + 3, and ∂M ∂N = −2x = , ∂y ∂x so the equation is exact. Thus, g (x, y ) = = M (x, y ) dx + h(y ) 3x2 − 2xy + 2 dx + h(y ) = x3 − x2 y + 2x + h(y ) Since ∂g ∂y = N (x, y ), we must have 6y 2 − x2 + 3 = ∂ x3 − x2 y + 2x + h(y ) = −x2 + h (y ) ∂y h (y ) = 6 y 2 + 3 Thus, h(y ) = 2y 3 + 3y + C , so g (x, y ) = x3 − x2 y + 2x + 2y 3 + 3y + C x3 − x2 y + 2 x + 2 y 3 + 3 y + C = 0 x3 − x2 y + 2 x + 2 y 3 + 3 y = k 21 k = −C (c) (ex sin y − 2y sin x) dx + (ex cos y + 2 cos x) dy = 0. Here M (x, y ) = ex sin y − 2y sin x and N (x, y ) = ex cos y + 2 cos x, and ∂M ∂N = ex cos y − 2 sin x = , ∂y ∂x so the equation is exact. Thus, g (x, y ) = = M (x, y ) dx + h(y ) ex sin y − 2y sin x dx + h(y ) = ex sin y + 2y cos x + h(y ) Since ∂g ∂y = N (x, y ), we must have ex cos y + 2 cos x = ∂x [e sin y + 2y cos x + h(y )] = ex cos y + 2 cos x + h (y ) ∂y h (y ) = 0 Thus, h(y ) = C , so g (x, y ) = ex sin y + 2y cos x + C ex sin y + 2y cos x + C = 0 x e sin y + 2y cos x = k 22 k = −C (d) √ x x2 +y 2 dx + ey + √ Here M (x, y ) = √ y x2 +y 2 x x2 +y 2 = 0. and N (x, y ) = ey + √ ∂M =− ∂y xy x2 + y 2 3 = y , x2 +y 2 and ∂N , ∂x so the equation is exact. Thus, g (x, y ) = M (x, y ) dx + h(y ) x = x2 ∂g ∂y ey + dx + h(y ) x2 + y 2 + h(y ) = Since + y2 = N (x, y ), we must have y x2 + y2 = ∂ ∂y x2 + y 2 + h(y ) = y x2 h (y ) = ey Thus, h(y ) = ey + C , so g (x, y ) = x2 + y 2 + ey + C x2 + y 2 + ey + C = 0 x2 + y 2 + ey = k 23 k = −C + y2 + h (y ) (e) ln y 1+x2 dx + arctan x y Here M (x, y ) = + y ln y dy = 0. ln y 1+x2 and N (x, y ) = arctan x y + y ln y, and ∂M 1 ∂N =− = , ∂y y (1 + x2 ) ∂x so the equation is exact. Thus, g (x, y ) = M (x, y ) dx + h(y ) ln y dx + h(y ) 1 + x2 = arctan x ln y + h(y ) = Since ∂g ∂y = N (x, y ), we must have ∂ arctan x arctan x + y ln y = [arctan x ln y + h(y )] = + h (y ) y ∂y y h (y ) = y ln y Thus, h(y ) = y ln y dy, which we solve by partial fractions, taking u = ln y du = so h(y ) = y2 ln y − 2 dv = y dy dy y v= y2 2 y2 y2 y dy = ln y − + C. 2 2 4 Thus, y2 y2 ln y − +C 2 4 y2 y2 arctan x ln y + ln y − +C =0 2 4 2 2 y y arctan x ln y + ln y − =k k = −C 2 4 g (x, y ) = arctan x ln y + 24 8. Find an integrating factor for the following differential equations and solve. (a) (x sin y ) dx + (2x cos y ) dy = 0. Here M (x, y ) = x sin y and N (x, y ) = 2x cos y , so ∂M = x cos y ∂y Since ∂M ∂y = ∂N ∂x , ∂N = 2 cos y. ∂x and the equation is not exact. However, 1 N ∂M ∂N − ∂y ∂x 11 − 2x = is a function of x alone, so we may solve by introducing an integrating factor 1 1 1 1 e2x e 2 − x dx = e 2 x−ln x = . x 1 Thus, if we take M1 (x, y ) = 1 e2x x N (x, y ) e2x x M (x, y ) 1 = e 2 x sin y and N1 (x, y ) = 1 = 2e 2 x cos y then M1 (x, y ) dx + N1 (x, y ) dy = 0 is exact: 1 ∂N1 ∂M1 = e 2 x cos y = . ∂y ∂x We may thus solve this differential equation for g (x, y ): g (x, y ) = M1 (x, y ) dx + h(y ) = e 2 x sin y dx + h(y ) 1 1 = 2e 2 x sin y + h(y ). Since ∂g ∂y = N1 (x, y ), we must have 1 2e 2 x cos y = 1 1 ∂ [2e 2 x sin y + h(y )] = 2e 2 x cos y + h (y ) ∂y h (y ) = 0 Thus, h(y ) = C , so 1 g (x, y ) = 2e 2 x sin y + C 1 2e 2 x sin y + C = 0 1 2e 2 x sin y = k, where k = −C . 25 (b) 1 x dx + x y dy = 0. Here M (x, y ) = 1 x and N (x, y ) = x , so y ∂M =0 ∂y Since ∂M ∂y = ∂N ∂x , and ∂N 1 =. ∂x y the equation is not exact. However, 1 N ∂M ∂N − ∂y ∂x =− 1 x is a function of x alone, so we may solve by introducing an integrating factor 1 1 e − x dx = e− ln x = . x x,y 1 Thus, if we take M1 (x, y ) = M (x ) = x2 and N1 (x, y ) = then M1 (x, y ) dx + N1 (x, y ) dy = 0 is exact: ∂M1 ∂N1 =0= . ∂y ∂x We may thus solve this differential equation for g (x, y ): g (x, y ) = M1 (x, y ) dx + h(y ) 1 dx + h(y ) x2 1 = − + h(y ). x = Since ∂g ∂y = N1 (x, y ), we must have 1 ∂ 1 = [− + h(y )] = h (y ) y ∂y x Thus, h(y ) = ln y + C , so g (x, y ) = − − 1 + ln y + C x 1 + ln y + C = 0 x 1 − + ln y = k, x where k = −C . 26 N (x,y ) x = 1 y (c) 2xy 2 dx + 3 x2 y + y dy = 0. 2 3 Here M (x, y ) = 2xy 2 and N (x, y ) = 2 x2 y + y , so ∂M = 4xy ∂y Since ∂M ∂y = ∂N ∂x , and ∂N = 3xy. ∂x the equation is not exact. However, 1 M ∂M ∂N − ∂y ∂x = 1 2y is a function of y alone, so we may solve by introducing an integrating factor 1 1 1 1 e− 2y dy = e− 2 ln y = y − 2 = √ . y 3 (x,y Thus, if we take M1 (x, y ) = M√y ) = 2xy 2 and N1 (x, y ) = √ √ 32 y + y then M1 (x, y ) dx + N1 (x, y ) dy = 0 is exact: 2x ∂M1 ∂N1 √ = 3x y = . ∂y ∂x We may thus solve this differential equation for g (x, y ): g (x, y ) = = M1 (x, y ) dx + h(y ) 3 2xy 2 dx + h(y ) 3 = x2 y 2 + h(y ). Since ∂g ∂y = N1 (x, y ), we must have 3 2√ ∂ 23 3√ √ x y+ y= [x y 2 + h(y )] = x2 y + h (y ) 2 ∂y 2 √ h (y ) = y 3 Thus, h(y ) = 2 y 2 + C , so 3 3 23 g (x, y ) = x2 y 2 + y 2 + C 3 3 23 x2 y 2 + y 2 + C = 0 3 3 23 x2 y 2 + y 2 = k, 3 where k = −C . 27 N (x,y ) √ y = (d) (x2 + 4xy ) dx + x dy = 0. Here M (x, y ) = x2 + 4xy and N (x, y ) = x, so ∂M = 4x ∂y Since ∂M ∂y = ∂N ∂x , and ∂N = 1. ∂x the equation is not exact. However, 1 N ∂M ∂N − ∂y ∂x =4− 1 x is a function of y alone, so we may solve by introducing an integrating factor 1 e4x . e 4− x dx = e4x−ln x = x 4x Thus, if we take M1 (x, y ) = ex M (x, y ) = xe4x +4ye4x and N1 (x, y ) = e4x 4x then M1 (x, y ) dx + N1 (x, y ) dy = 0 is exact: x N (x, y ) = e ∂M1 ∂N1 = 4e4x = . ∂y ∂x We may thus solve this differential equation for g (x, y ): g (x, y ) = M1 (x, y ) dx + h(y ) xe4x + 4ye4x dx + h(y ) = = Since ∂g ∂y 1 1 4x xe − e4x + ye4x + h(y ). 4 16 = N1 (x, y ), we must have e4x = 1 ∂ 1 4x [ xe − e4x + ye4x + h(y )] = e4x + h (y ) ∂y 4 16 h (y ) = 0 Thus, h(y ) = C , so 1 4x 1 xe − e4x + ye4x + C 4 16 1 4x 1 xe − e4x + ye4x + C = 0 4 16 1 4x 1 xe − e4x + ye4x = k, 4 16 g (x, y ) = where k = −C . 28 (e) xy dx + (y 2 + x2 ) dy = 0. Here M (x, y ) = xy and N (x, y ) = y 2 + x2 , so ∂M =x ∂y Since ∂M ∂y = ∂N ∂x , and ∂N = 2x. ∂x the equation is not exact. However, 1 M ∂M ∂N − ∂y ∂x =− 1 y is a function of y alone, so we may solve by introducing an integrating factor 1 e y dy = eln y = y Thus, if we take M1 (x, y ) = yM (x, y ) = xy 2 and N1 (x, y ) = yN (x, y ) = y 3 + x2 y then M1 (x, y ) dx + N1 (x, y ) dy = 0 is exact: ∂M1 ∂N1 = 2xy = . ∂y ∂x We may thus solve this differential equation for g (x, y ): g (x, y ) = M1 (x, y ) dx + h(y ) xy 2 dx + h(y ) = = Since ∂g ∂y 122 x y + h(y ). 2 = N1 (x, y ), we must have y 3 + x2 y = ∂ 122 [ x y + h(y )] = x2 y + h (y ) ∂y 2 h (y ) = y 3 1 Thus, h(y ) = 4 y 4 + C , so 122 14 x y + y +C 2 4 122 14 x y + y +C =0 2 4 122 14 x y + y = k, 2 4 g (x, y ) = where k = −C . 29 9. Make a substitution to solve the following differential equations. (a) y = (2x + y )2 − 6, y (0) = 4. Let v = 2x + y , so v = 2 + y and y = v − 2. Then, the equation becomes: v − 2 = v2 − 6 v = v2 − 4 This equation is separable: dv = 1 dx v2 − 4 dv = −4 v2 1 dx = x + C We can do this integral by partial fractions: 1 A B (A + B )v + (−2A + 2B ) = + = , v2 − 4 v+2 v−2 v2 − 4 so A+B =0 −2A + 2B = 1 1 This pair of linear equations had solution A = − 4 and B = 1 : 4 dv 1 1 1 = − dv −4 4 v−2 v+2 1 = ln |v − 2| − ln |v + 2| 4 v−2 1 |. = ln | 4 v+2 v2 30 Thus, we have 1 v−2 ln | |=x+C 4 v+2 v−2 ln | | = 4x + C v+2 v−2 | | = De4x v+2 v−2 = ±De4x v+2 4 = ±De4x 1− v+2 4 −2 v= 1 De4x 4 y= − 2 − 2x 1 De4x Lastly, we use the initial condition y (0) = 4 to solve for the constant D: 4 = y (0) = 4 1 D 1 D=− 3 1 D=± , 3 −2 so, in either case, y= 4 − 2 − 2x. 1 − 1 e4x 3 (b) y = e2x+y + 1. Let v = 2x + y , so v = 2 + y . Then, the equation becomes v − 2 = y = ev + 1 v = ev + 3 This equation is separable: dv = 1 dx +3 ev dv = ev + 3 31 1 dx = x + C Let u = ev , so du = ev dv and du u = dv . Then, dv = ev + 3 du u(u + 3) 11 1 1 − = 3u 3u+3 1 1 = ln |u| − ln |u + 3| 3 3 u 1 | = ln | 3 u+3 Thus, 1 u ln | |=x+C 3 u+3 u ln | | = 3x + C u+3 u | | = De3x u+3 u = ±De3x u+3 3 1− = ±De3x u+3 3 −3 u= 1 ± De3x Thus, 3 −3 1 ± De3x 3 v = ln −3 , 1 ± De3x ev = and 2x + y = ln y = ln 3 −3 1 ± De3x 3 − 3 − 2x 1 ± De3x 32 0.2 Higher Order Differential Equations 1. Find the general solution for the following linear differential equations. (a) y − y − 2y = 0. This has characteristic polynomial λ2 − λ − 2 = (λ + 1)(λ − 2), so it has two distinct real roots, λ1 = −1 and λ2 = 2, so that y = c1 e−x + c2 e2x . (b) 4y + 20y + 25y = 0. This has characteristic polynomial 4λ2 + 20λ + 25 = (2λ + 5)2 , so it 5 has a single real root λ = − 2 . Thus, 5 y = (c1 + c2 x)e− 2 x . (c) y − 4y + 5y = 0. This has characteristic polynomial λ2 − 4λ + 5 = 0, which has roots λ2 = 2 − i λ1 = 2 + i, so y = e2x (c1 cos x + c2 sin x). (d) 12y + 16y + 5y = 0.. This has characteristic polynomial 12λ2 + 16λ + 5 = (2λ + 1)(6λ + 5), 1 5 so it has two distinct real roots, λ1 = − 2 and λ2 = − 6 , so that 1 5 y = c1 e − 2 x + c2 e − 6 x . (e) y + 3y + 5y = 0.. This has characteristic polynomial λ2 + 3λ + 5 = 0, which has roots √ √ 3 11 3 11 λ1 = − + i , λ2 = − − i 2 2 2 2 so √ y=e 3 −2x √ 11 11 (c1 cos x + c2 sin x). 2 2 33 (f) y − 3y − 2y = 0. This has characteristic polynomial λ3 − 3λ − 2 = 0. We make a guess for the first root, taking λ1 = −1: (−1)3 − 3(−3) − 2(−1) = 0. Using long division, you can show λ3 − 3λ − 2 = (λ + 1)(λ2 − λ − 2) = (λ + 1)(λ + 1)(λ − 2) so the other roots are λ2 = −1 and λ3 = 2. Thus, the general solution is y = c1 e−x + c2 xe−x + c3 e2x . (g) y − 6y + 12y − 8y = 0. This has characteristic polynomial λ3 − 6λ2 + 12λ − 8 = 0. We make a guess for the first root, taking λ1 = 2: (2)3 − 6(2)2 + 12(2) − 8 = 0. Using long division, you can show λ3 − 6λ2 + 12λ − 8 = (λ − 2)(λ2 − 4λ + 4) = (λ − 2)3 so the other roots are λ2 = 2 and λ3 = 2. Thus, the general solution is y = c1 e2x + c2 xe2x + c3 x2 e2x . (h) y − 7y + 17y − 15y = 0. This has characteristic polynomial λ3 − 7λ2 + 17λ − 15 = 0. We make a guess for the first root, taking λ1 = 3: (3)3 − 7(3)2 + 17(3) − 15 = 0. Using long division, you can show λ3 − 7λ2 + 17λ − 15 = (λ − 3)(λ2 − 4λ + 5) Using the quadratic formula, the other roots are λ2 = 2 + i and λ3 = 2 − i. Thus, the general solution is y = c1 e3x + c2 e2x cos x + c3 e2x sin x. 34 (i) y (6) − y = 0. This has characteristic polynomial λ6 − 1 = (λ3 − 1)(λ3 + 1) = (λ − 1)(λ2 + λ + 1)(λ + 1)(λ2 − λ + 1) Thus λ1 = 1 and λ2 = −1 are roots. The roots of the two remaining quadratic polynomials can be found using the quadratic formula: √ √ 3 1 3 1 and λ3 , λ4 = ± i . λ3 , λ4 = − ± i 2 2 2 2 The general solution is thus √ x −x y = c1 e + c2 e + c3 e −1 2 √ 3 x 2 cos + c4 e 1 −2 sin √ 1 2 + c5 e cos 3 x 2 1 3 x 2 √ + c6 e 2 sin 3 x. 2 (j) y (4) − 8y + 16y = 0. This has characteristic polynomial λ4 − 8λ2 + 16 = (λ2 − 4)2 = (λ + 2)2 (λ − 2)2 Thus, the roots are λ1 = λ2 = −2 and λ3 = λ4 = 2. The general solution is y = c1 e−2x + c2 xe−2x + c3 e2x + c4 xe2x . 2. Solve y − 4y = 0, subject to y (0) = 1, y (0) = −1. Start by finding the general solution. This has characteristic polynomial λ2 − 4 = (λ + 2)(λ − 2), so it has two distinct real roots, λ1 = −2 and λ2 = 2, so that y = c1 e−2x + c2 e2x . Now, y = −2c1 e−2x + 2c2 e2x , so, imposing initial conditions, we have 1 = y (0) = c1 + c2 −1 = y (0) = −2c1 + 2c2 , giving two linear equations for the two constants c1 and c2 . Adding two 1 times the first to the second gives 1 = 4c2 , so c2 = 4 . From the first, we 3 see that c1 = 4 . Thus, 3 1 y = e−2x + e2x . 4 4 35 3. Find the general solution to the following non-homogeneous equations with constant coefficients, using the method of undetermined coefficients. (a) y − 3y − 4y = 5x2 . We start by finding all solutions to the homogeneous equation y − 3y − 4y = 0. This has characteristic equation λ2 − 3λ − 4 = (λ + 1)(λ − 4) = 0, which has distinct real roots λ1 = −1 and λ2 = 4, so the solution to the homogeneous equation is yh = c1 e−x + c2 e4x , where c1 and c2 are arbitrary constants. We next look for a particular solution by the method of undetermined coefficients. We look for a solution of the form yp = A2 x2 + A1 x + A0 , so yp = 2A2 x + A1 yp = 2A2 Substituting these into y − 3y − 4y = 5x2 , we get (2A2 ) − 3(2A2 x + A1 ) − 4(A2 x2 + A1 x + A0 ) = 5x2 Collecting like terms on the right hand side, we get −4A2 x2 + (−6A2 − 4A1 )x + (−3A1 − 4A0 ) = 5x2 So, equating the coefficients of like terms on the right and left hand sides: −4A2 = 5 −6A2 − 4A1 = 0 −3A1 − 4A0 = 0 95 This system of linear equations has solution A0 = − 32 , A1 = 5 A2 = − 4 . Thus, 5 15 95 yp = − x2 + x + − 4 8 32 and 5 15 95 y = yh + yp = c1 e−x + c2 e4x − x2 + x + − . 4 8 32 36 15 8, and (b) y − 3y − 18y = 4e2x . We start by finding all solutions to the homogeneous equation y − 3y − 18y = 0. This has characteristic equation λ2 − 3λ − 18 = (λ + 3)(λ − 6) = 0, which has distinct real roots λ1 = −3 and λ2 = 6, so the solution to the homogeneous equation is yh = c1 e−3x + c2 e6x , where c1 and c2 are arbitrary constants. We next look for a particular solution by the method of undetermined coefficients. We look for a solution of the form yp = Ae2x , so yp = 2Ae2x yp = 4Ae2x Substituting these into y − 3y − 18y = 4e2x , we get (4Ae2x ) − 3(2Ae2x ) − 18(Ae2x ) = 4e2x Collecting like terms on the right hand side, we get −20Ae2x = 4e2x So, equating the coefficients of like terms on the right and left hand sides: −20A = 4 1 1 Thus, A = − 5 , yp = − 5 e2x , and 1 y = yh + yp = c1 e−3x + c2 e6x − e2x . 5 37 (c) y − 5y = 2 sin 2x + 5 cos 2x. We start by finding all solutions to the homogeneous equation y − 5y = 0. This has characteristic equation λ2 − 5λ = λ(λ − 5) = 0, which has distinct real roots λ1 = 0 and λ2 = 5, so the solution to the homogeneous equation is yh = c1 + c2 e5x , where c1 and c2 are arbitrary constants. We next look for a particular solution by the method of undetermined coefficients. We look for a solution of the form yp = A sin 2x + B cos 2x, so yp = 2A cos 2x − 2B sin 2x yp = −4A sin 2x − 4B cos 2x Substituting these into y − 5y = 2 sin 2x + 5 cos 2x, we get (−4A sin 2x − 4B cos 2x) − 5(2A cos 2x − 2B sin 2x) = 2 sin 2x +5 cos 2x Collecting like terms on the right hand side, we get (−4A + 10B ) sin 2x + (−4B − 10A) cos 2x = 2 sin 2x + 5 cos 2x So, equating the coefficients of like terms on the right and left hand sides: −4A + 10B = 2 −4B − 10A = 5 1 This system of linear equations has solution A = − 2 and B = 0. Thus, 1 yp = − sin 2x 2 and 1 y = yh + yp = c1 + c2 e5x − sin 2x. 2 38 (d) y + 2y = 2e2x − 3 sin 3x + 3 cos 3x. We start by finding all solutions to the homogeneous equation y + 2y = 0. This has characteristic equation λ2 + 2λ = λ(λ + 2) = 0, which has distinct real roots λ1 = −2 and λ2 = 0, so the solution to the homogeneous equation is yh = c1 + c2 e−2x , where c1 and c2 are arbitrary constants. We next look for a particular solution by the method of undetermined coefficients. We look for a solution of the form yp = Ae2x + B sin 3x + C cos 3x, so yp = 2Ae2x + 3B cos 3x − 3C sin 3x yp = 4Ae2x − 9B sin 3x − 9C cos 3x Substituting these into y + 2y = 2e2x − 3 sin 3x + 3 cos 3x, we get (4Ae2x − 9B sin 3x − 9C cos 3x) + 2(2Ae2x + 3B cos 3x − 3C sin 3x) = 2e2x − 3 sin 3x + 3 cos 3x. Collecting like terms on the right hand side, we get 8Ae2x +(−9B −6C ) sin 3x+(6B −9C ) cos 3x = 2e2x −3 sin 3x+3 cos 3x So, equating the coefficients of like terms on the right and left hand sides: 8A = 2 −9B − 6C = 3 6B − 9C = 3 This system of linear equations has solution A = 1 C = − 13 . Thus, yp = 1 4, B= 5 13 , 1 2x 5 1 e+ sin 3x − cos 3x 4 13 13 and 1 5 1 y = yh + yp = c1 + c2 e−2x + e2x + sin 3x − cos 3x. 4 13 13 39 and (e) y − 2y − 15y = 32x2 ex + 40xex . We start by finding all solutions to the homogeneous equation y − 2y − 15y = 0. This has characteristic equation λ2 − 2λ − 15 = (λ + 3)(λ − 5) = 0, which has distinct real roots λ1 = −3 and λ2 = 5, so the solution to the homogeneous equation is yh = c1 e−3x + c2 e5x , where c1 and c2 are arbitrary constants. We next look for a particular solution by the method of undetermined coefficients. We look for a solution of the form yp = A2 x2 ex + A1 xex + A0 ex , so yp = A2 x2 ex + 2A2 xex + A1 xex + A1 ex + A0 ex yp = A2 x2 ex + 4A2 xex + 2A2 ex + A1 xex + 2A1 ex + A0 ex Substituting these into y − 2y − 15y = 32x2 ex + 40xex , we get (A2 x2 ex + 4A2 xex + 2A2 ex + A1 xex + 2A1 ex + A0 ex ) − 2(A2 x2 ex + 2A2 xex + A1 xex + A1 ex + A0 ex ) − 15(A2 x2 ex + A1 xex + A0 ex ) = 32x2 ex + 40xex Collecting like terms on the right hand side, we get −16A2 x2 ex − 16A1 xex + (−16A0 + 2A2 )ex = 32x2 ex + 40xex So, equating the coefficients of like terms on the right and left hand sides: −16A2 = 32 −16A1 = 40 −16A0 + 2A2 = 0 5 1 This system of linear equations has solution A0 = − 4 , A1 = − 2 , and A2 = −2. Thus, 5 1 yp = −2x2 ex − xex − ex 2 4 and 1 y = yh + yp = c1 e−3x + c2 e5x − 2x2 ex + ex . 4 40 (f) y − 7y + 17y − 15y = 15x2 − 60x + 27. We start by finding all solutions to the homogeneous equation y − 7y − 17y − 15y = 0. This has characteristic equation λ3 − 7λ2 + 17λ − 15 = (λ − 3)(λ2 − 4λ + 5) = 0. Thus, λ1 = 3 is one root of this equation. We find the other two roots using the quadratic formula: λ2 , λ3 = 2 ± i. The solution to the homogeneous equation is then yh = c1 e3x + c2 e2x cos x + c3 e2x sin x, where c1 , c2 and c3 are arbitrary constants. We next look for a particular solution by the method of undetermined coefficients. We look for a solution of the form yp = A2 x2 + A1 x + A0 , so yp = 2A2 x + A1 yp = 2A2 yp = 0 Substituting these into y − 7y + 17y − 15y = 15x2 − 60x + 27, we get −7(2A2 ) + 17(2A2 x + A1 ) − 15(A2 x2 + A1 x + A0 ) = 15x2 − 60x + 27 Collecting like terms on the right hand side, we get −15A2 x2 + (34A2 − 15A1 )x + (17A1 − 15A0 ) = 15x2 − 60x + 27 So, equating the coefficients of like terms on the right and left hand sides: −15A2 = 15 34A2 − 15A1 = −60 17A1 − 15A0 = 27 This system of linear equations has solution A0 = A2 = −1. Thus, 26 37 yp = −x2 + x + 15 225 and 37 225 , y = yh + yp = c1 e3x + c2 e2x cos x + c3 e2x sin x − x2 + 41 A1 = 26 15 , and 26 37 x+ . 15 225 (g) y (4) + 3y + 2y = 144e2x . We start by finding all solutions to the homogeneous equation y (4) + 3y + 2y = 0. This has characteristic equation λ4 + 3λ3 + 2λ2 = λ2 (λ + 1)(λ + 2) = 0, which has a repeated root, λ1 = λ2 = 0, and two distinct roots, λ3 = −1 and λ4 = −2. The solution to the homogeneous equation is then yh = c1 + c2 x + c3 e−x + c4 e−2x , where c1 , c2 , c3 and c4 are arbitrary constants. We next look for a particular solution by the method of undetermined coefficients. We look for a solution of the form yp = Ae2x , so yp = 2Ae2x yp = 4Ae2x yp = 8Ae2x (4) yp = 16Ae2x Substituting these into y (4) + 3y + 2y = 10e2x , we get (16Ae2x ) + 3(8Ae2x ) + 2(4Ae2x ) = 144e2x Collecting like terms on the right hand side, we get 48Ae2x = 144e2x So, A = 3, yp = 3e2x and y = yh + yp = c1 + c2 x + c3 e−x + c4 e−2x + 3e2x 42 4. Solve the following Cauchy-Euler equations. (a) x2 y − 5xy + 13y = 0. Here, a = −5 and b = 13, so the auxiliary equation is λ2 − 6λ +13 = 0. Using the quadratic formula, we see this has roots λ1 , λ2 = 3 ± 2i, so the solution to the equation is y = c1 x3 cos (2 ln x) + c2 x3 sin (2 ln x), where c1 and c2 are arbitrary constants. (b) x2 y − xy + 5y = 0. Here, a = −1 and b = 5, so the auxiliary equation is λ2 − 2λ + 5 = 0. Using the quadratic formula, we see this has roots λ1 , λ2 = 1 ± 2i, so the solution to the equation is y = c1 x cos (2 ln x) + c2 x sin (2 ln x), where c1 and c2 are arbitrary constants. (c) x2 y + 6xy + 25y = 0. Here, a = 6 and b = 25, so the auxiliary equation is λ2 + 5λ + 25 = 0. Using the quadratic formula, we see this has roots √ 5 λ1 , λ2 = − ± i5 3, 2 so the solution to the equation is √ √ 5 5 y = c1 x− 2 cos (5 3 ln x) + c2 x− 2 sin (5 3 ln x), where c1 and c2 are arbitrary constants. 43 5. Find the general solution of the following non-homogeneous Cauchy-Euler equations by the method of variation of parameters. (a) x2 y − 2xy + 2y = x3 ex . We start by finding all solutions to the homogeneous equation, x2 y − 2xy + 2y = 0. This is a Cauchy-Euler equation with a = −2 and b = 2 and auxiliary equation λ2 − 3λ + 2 = (λ − 1)(λ − 2) = 0. This has two distinct real roots, λ1 = 1 and λ2 = 2, so the solution to the homogeneous equation is yh = c1 x + c2 x2 , where c1 and c2 are arbitrary constants. Next, we compute the Wronskian of the fundamental solutions y1 = x and y2 = x2 : x x2 = x2 . W [y1 , y2 ] = 1 2x The method of variation of parameters tells us that the particular solution is of the form xp = v1 y1 + v2 y2 , where x2 (x3 ex ) dx x2 ( x2 ) v1 (x) = − xex dx =− = −xex + ex and x(x3 ex ) dx x2 (x2 ) v2 (x) = = ex dx = ex . Thus, yp = x(−xex + ex ) + x2 (ex ) = xex and y = yh + yp = c1 x + c2 x2 − xex . 44 (b) x2 y − 7xy + 15y = x6 cos x. We start by finding all solutions to the homogeneous equation, x2 y − 7xy + 15y = 0. This is a Cauchy-Euler equation with a = −7 and b = 15 and auxiliary equation λ2 − 8λ + 15 = (λ − 3)(λ − 5) = 0. This has two distinct real roots, λ1 = 3 and λ2 = 5, so the solution to the homogeneous equation is yh = c1 x3 + c2 x5 , where c1 and c2 are arbitrary constants. Next, we compute the Wronskian of the fundamental solutions y1 = x3 and y2 = x5 : W [y1 , y2 ] = x3 3x2 x5 = 2x7 . 5x4 The method of variation of parameters tells us that the particular solution is of the form xp = v1 y1 + v2 y2 , where x5 (x6 cos x) dx x2 (2x7 ) v1 (x) = − x2 cos x dx =− = −x2 sin x + 2 sin x − 2x cos x and v2 (x) = = x3 (x6 cos x) dx x2 (2x7 ) 2 cos x dx = 2 sin x. Thus, yp = x3 (−x2 sin x+2 sin x−2x cos x)+x2 (2 sin x) = x5 sin x+2x3 sin x−2x4 cos x and y = yh + yp = c1 x3 + c2 x5 + x5 sin x + 2x3 sin x − 2x4 cos x. 45 (c) x2 y − 3xy + 4y = x5 We start by finding all solutions to the homogeneous equation, x2 y − 3xy + 4y = 0. This is a Cauchy-Euler equation with a = −3 and b = 4 and auxiliary equation λ2 − 4λ + 4 = (λ − 2)2 = 0. This has two equal real roots, λ1 = λ2 = 2, so the solution to the homogeneous equation is yh = c1 x2 + c2 x2 ln x, where c1 and c2 are arbitrary constants. Next, we compute the Wronskian of the fundamental solutions y1 = x2 and y2 = x2 ln x: W [y1 , y2 ] = x2 x2 ln x = x3 . 2x 2x ln x + x The method of variation of parameters tells us that the particular solution is of the form xp = v1 y1 + v2 y2 , where v1 (x) = − =− x2 ln x(x5 ) dx x2 ( x3 ) x2 ln x dx 1 1 = − x3 ln x + x3 3 9 and v2 (x) = = = x2 (x5 ) dx x2 (x3 ) x2 dx 13 x. 3 Thus, 1 1 1 1 4 yp = x2 (− x3 ln x + x3 ) + x2 ln x( x3 ) = − x5 ln x + x5 3 9 3 3 9 and 4 1 y = yh + yp = c1 x2 + c2 x2 ln x − x5 ln x + x5 . 3 9 46 (d) x2 y − 5xy + 9y = x3 ln x We start by finding all solutions to the homogeneous equation, x2 y − 5xy + 9y = 0. This is a Cauchy-Euler equation with a = −5 and b = 9 and auxiliary equation λ2 − 6λ + 9 = (λ − 3)2 = 0. This has two equal real roots, λ1 = λ2 = 3, so the solution to the homogeneous equation is yh = c1 x3 + c2 x3 ln x, where c1 and c2 are arbitrary constants. Next, we compute the Wronskian of the fundamental solutions y1 = x3 and y2 = x3 ln x: W [y1 , y2 ] = x3 3x2 x3 ln x = x5 . 3x ln x + x2 2 The method of variation of parameters tells us that the particular solution is of the form xp = v1 y1 + v2 y2 , where v1 ( x ) = − =− x3 ln x(x3 ln x) dx x2 ( x5 ) (ln x)2 dx x 1 = − (ln x)3 3 and v2 ( x ) = = = x3 (x3 ln x) dx x2 (x5 ) ln x dx x 1 (ln x)2 . 2 Thus, 1 1 1 yp = x3 (− (ln x)3 ) + x3 ln x( (ln x)2 ) = (ln x)3 3 2 6 and 1 y = yh + yp = c1 x3 + c2 x3 ln x + (ln x)3 . 6 47 (e) x2 y + 7xy + 9y = x4 We start by finding all solutions to the homogeneous equation, x2 y + 7xy + 9y = 0. This is a Cauchy-Euler equation with a = 7 and b = 9 and auxiliary equation λ2 + 6λ + 9 = (λ + 3)2 = 0. This has two equal real roots, λ1 = λ2 = −3, so the solution to the homogeneous equation is yh = c1 x−3 + c2 x−3 ln x, where c1 and c2 are arbitrary constants. Next, we compute the Wronskian of the fundamental solutions y1 = x−3 and y2 = x−3 ln x: W [y1 , y2 ] = x −3 −3x−4 x−3 ln x = x−7 . −3x ln x + x−4 −4 The method of variation of parameters tells us that the particular solution is of the form xp = v1 y1 + v2 y2 , where v1 ( x ) = − =− x−3 ln x(x4 ) dx x2 (x−7 ) x6 ln x dx 1 1 = − x7 ln x + x7 7 49 and v2 ( x ) = = = x−3 (x4 ) dx x2 (x−7 ) x6 dx 17 x. 7 Thus, 1 1 1 14 yp = x−3 (− x7 ln x + x7 ) + x−3 ln x( x7 ) = x 7 49 7 49 and y = yh + yp = c1 x−3 + c2 x−3 ln x + 48 14 x. 49 6. Solve the following linear equations with constant coefficients by the method of variation of parameters. 2x (a) y − 4y + 4y = ex . We start by finding all solutions to the homogeneous equation y − 4y + 4y = 0. This has characteristic equation λ2 − 4λ + 4 = (λ − 2)2 = 0, which has repeated roots λ1 = λ2 = 2, so the solution to the homogeneous equation is yh = c1 e2x + c2 xe2x , where c1 and c2 are arbitrary constants. Next, we compute the Wronskian of the fundamental solutions y1 = e2x and y2 = xe2x : W [y1 , y2 ] = e2x 2e2x e 2x xe2x = e4x . + 2xe2x The method of variation of parameters tells us that the particular solution is of the form xp = v1 y1 + v2 y2 , where e2x x xe2x v1 ( x ) = − =− e4x dx 1 dx = −x and e2x x e2x v2 (x) = e4x dx 1 dx x = ln x = Thus, yp = −xe2x + x(ln x)e2x and y = yh + yp = c1 e2x + c2 xe2x − xe2x + x(ln x)e2x . 49 (b) y − 3y − 18y = 27e6x . We start by finding all solutions to the homogeneous equation y − 3y − 18y = 0. This has characteristic equation λ2 − 3λ − 18 = (λ + 3)(λ − 6) = 0, which has distinct real roots λ1 = −3 and λ2 = 6, so the solution to the homogeneous equation is yh = c1 e−3x + c2 e6x , where c1 and c2 are arbitrary constants. Next, we compute the Wronskian of the fundamental solutions y1 = e−3x and y2 = e6x : W [y1 , y2 ] = e−3x −3e−3x e6x = 9e3x . 6e6x The method of variation of parameters tells us that the particular solution is of the form xp = v1 y1 + v2 y2 , where e6x 27e6x dx 9e3x v1 ( x ) = − 3e3x dx =− =− e9x 3 and e−3x 27e6x dx 9e3x v2 ( x ) = = 3 dx = 3x Thus, yp = − e6x e9x −3x e + 3xe6x = − + 3xe6x 3 3 and y = yh + yp = c1 e−3x + c2 e6x − 50 e6x + 3xe6x = c1 e−3x + c2 e6x + 3xe6x . 3 7. Suppose that the solutions to y − 6y + 12y − 8y = 0 are e2x , xe2x , and x2 e2x . Show that these functions are linearly independent. To show they are linearlly independent, we must show that their Wronskian is a non-zero function. e2x W [e , xe , x e ] = 2e2x 4e2x 2x 2x xe2x e + 2xe2x 4e2x + 4xe2x 2 2x 2x x2 e2x 2xe + 2x2 e2x = 2e6x = 0. 2x 2x 2 2x 2e + 8xe + 4x e 2x Therefore, they are linearly independent. 8. Suppose that the solutions to y − 4y + 5y = 0 are e2x cos x and e2x sin x. Show that these functions are linearly independent. To show they are linearlly independent, we must show that their Wronskian is a non-zero function. W [e2x cos x, e2x sin x] = 2e 2x e2x cos x cos x − e2x sin x =e 4x 2 cos x + e =e 4x 4x = 0. Therefore, they are linearly independent. 51 2 sin x e 2x e2x sin x cos x + 2e2x sin x 9. ex − 1 is a solution to (ex + 1)y − 2y − ex y = 0. Find another linearly independent solution. Let y1 = 3x − 1, sp y1 = y1 = ex . Start by checking that y1 is a solution of the differential equation: (ex + 1)y1 − 2y1 − ex y1 = (ex + 1)ex − 2e− ex (ex − 1) = e2x + ex − 2ex − e2x + ex = 0. Let the unknown linearly independent solution be y2 . Then, we must have W [y1 , y2 ] = y1 y1 y2 y2 = y1 y2 − y2 y1 = (ex − 1)y2 − ex y2 . On the other hand, applying Abel’s formula to the differential equation (ex + 1)y − 2y − ex y = 0, W [y1 , y2 ] = Ce 2 ex +1 dx du u−1 Let u = ex + 1, so du = ex dx = (u − 1) dx and 2 dx = ex + 1 = dx: 2 du u(u − 1) 2 2 − du = u u−1 = −2 ln u + 2 ln u − 1 = ln = ln so W [y1 , y2 ] = C ex ex +1 u−1 u ex ex + 1 2 . 52 2 2 Setting these two expressions for the Wronskian equal, we get a linear differential equation for y2 : 2 ex ex + 1 x e2x e y2 − x y2 = C x . e −1 (e + 1)2 (ex − 1) (ex − 1)y2 − ex y2 = C x 2x e e Here, P (x) = − ex −1 and Q(x) = C (ex +1)2 (ex −1) so ex e2x e− ex −1 dx + C2 x + 1)2 (ex − 1) (e x x e2x = eln (e −1) C e− ln (e −1) dx + C2 x + 1)2 (ex − 1) (e e2x = (ex − 1) C dx + C2 x + 1)2 (ex − 1)2 (e e2x = (ex − 1) C dx + C2 2x − 1)2 (e y2 = e ex ex −1 C Let u = e2x − 1, so du = 2e2x dx and du 2 = e2x dx. Then, e2x dx = (e2x − 1)2 1 du 2u2 1 =− 2u 1 1 = − 2x , 2e −1 so y2 = (ex − 1) − = C1 C 1 + C2 2 e2x − 1 ex − 1 + C2 (ex − 1), e2x − 1 where C1 = − C . 2 Thus, e x −1 e2x −1 is another linearly independent solution. 53 10. Solve the following differential equations by reduction of order. (a) y + x(y )2 = 0. Let v = y , so v = y and the differential equation becomes v +xv 2 = 0. This is a separable equation: dv = −xv 2 dx dv = −x dx v2 dv = − x dx v2 1 x2 − = − + C1 v 2 1 v = x2 2 + C1 Lastly, v = y , so y= v dx = = Let u = √x , 2C1 so du = x2 2 1 C1 √1 2C1 and 1 dx + C1 1 2 dx 1 + 2x 1 C √ 2C1 du = dx: 1 1 2 dx C1 1 + 2x 1 C √ 2C1 1 = du C1 1 + u2 2 = arctan u + C2 C1 2 x = arctan √ + C2 . C1 2C1 y= 54 (b) y y = 2. Let v = y , so v = y and the differential equation becomes v v = 2. This is separable: dv =2 dx v dv = 2 dx v v dv = 2 dx v2 = 2x + C1 2 v = ± 4x + C1 Lastly, v = y , so y= v dx =± 4x + C1 dx 3 1 = ± (4x + C1 ) 2 = C2 . 6 55 0.3 The Laplace Transform 1. Find the following Laplace transforms of the following functions using the definition. (a) f (t) = 1. The Laplace Transform of f (t) = 1 is ∞ e−st dt F (s) = 0 R e−st dt = lim R→∞ 0 R 1 − e−st R→∞ s 0 1 −sR 1−e = lim R→∞ s 1 = s = lim if s > 0 and diverges otherwise. (b) f (t) = t2 . The Laplace Transform of f (t) = t2 is ∞ t2 e−st dt F (s) = 0 R t2 e−st dt = lim R→∞ = lim R→∞ 0 1 − te−st s R R + 0 0 1 2 = lim − Re−sR + R→∞ s s 1 = lim − Re−sR − R→∞ s 1 = lim − Re−sR − R→∞ s 2 =3 s 1 −st 2te dt s 1 − te−st s 2 Re−sR + s2 2 Re−sR + s2 R + 0 1 s R e−st dt 0 R 2 1 − e−st s2 s 0 2 −sR 1−e s3 if s > 0 and diverges otherwise. 56 (c) f (t) = eat . The Laplace Transform of f (t) = eat is ∞ eat e−st dt F (s) = 0 R e−(s−a)t dt = lim R→∞ 0 R 1 −(s−a)t e R→∞ s−a 0 1 1 − e−(s−a)R = lim − R→∞ s−a 1 = s−a = lim − and diverges otherwise. 57 if s > a (d) f (t) = cos at. The Laplace Transform of f (t) = cos at is ∞ e−st cos at dt F (s) = 0 R e−st cos at dt = lim R→∞ We compute R 0 R −st e 0 0 cos at dt by parts: R 1 a e−st cos at dt = − e−st cos at − s s 0 1 1 −sR =−e cos aR s s − = a s 1 − e−st sin at s R e−st sin at dt 0 R + 0 a s R e−st cos at dt 0 1 a a2 1 − e−sR cos aR + 2 e−sR sin aR − 2 s s s s R e−st cos at dt 0 Thus, 1+ R a2 s2 −st e R e−st cos at dt = 0 cos at dt = 0 1 1+ a2 s2 1 a 1 − e−sR cos aR + 2 e−sR sin aR s s s 1 1 a − e−sR cos aR + 2 e−sR sin aR s s s so R e−st cos at dt F (s) = lim R→∞ = 0 1 s if s > 0 2 1 + a2 s s =2 s + a2 if s > 0 and diverges otherwise. 58 2. Use the properties of the Laplace transform and the table at the front of the book to determine the Laplace Transform of the following functions. (a) 5 + 11t. L {5 + 11t} = 5L {1} + 11L {t} = 5 11 + 2. s s (b) t4 e−2t . L t4 e−2t = 4! 24 = . 5 (s + 2) (s + 2)5 (c) t3 sin 3t. d3 L {sin 3t} ds3 3 d3 = (−1)3 3 2 ds s + 32 72s(s2 − 9) = (s2 + 9)4 L t3 sin 3t = (−1)3 (d) e5t sin 4t. L e5t sin 4t = 4 4 =2 . 2 + 42 (s − 5) s − 10s + 41 (e) t2 cos 2t + t3 e−4t . L t2 cos 2t + t3 e−4t = L t2 cos 2t + L t3 e−4t d2 3! L {cos 2t} + ds2 (s + 4)4 d2 s 6 = 22 + ds s + 22 (s + 4)4 2s(s2 − 12) 6 = + (s2 + 4)3 (s + 4)4 = (−1)2 59 3. Find the Laplace transform of the following step functions. (a) g (t) = 0 (t − 3)4 t<3 . t≥3 This is of the form u3 (t)f (t − 3) for f (t) = t3 . Thus L {g (t)} = L {u4 (t)f (t − 3)} = e−3s L {f (t)} = e−3s L t3 = e−4s 3!s4 . (b) g (t) = 0 t−5 t<5 . t≥5 This is of the form u5 (t)f (t − 5) for f (t) = t. Thus L {g (t)} = L {u5 (t)f (t − 5)} = e−5s L {f (t)} = e−5s L {t} = e−4s (c) g (t) = 0 t−1 1 . s2 t<6 . t≥6 Here, g (t) = 0 (t − 6) + 5 t<6 , t≥6 so g (t) is of the form u6 (t)f (t − 6) for f (t) = t + 5. Thus L {g (t)} = L {u6 (t)f (t − 6)} = e−6s L {f (t)} = e−6s L {t + 5} = e−4s (d) g (t) = 0 t2 1 5 + s2 s . t<2 . t≥2 Here, g (t) = 0 ((t − 2) + 2)2 t<2 , t≥2 so g (t) is of the form u2 (t)f (t − 2) for f (t) = (t + 2)2 = t2 + 4t + 4. Thus L {g (t)} = L {u2 (t)f (t − 2)} = e−2s L {f (t)} = e−2s L t2 + 4t + 4 = e−2s 60 2 4 4 + 2+ s3 s s . (e) g (t) = 0 t3 + 1 t<1 . t≥1 Here, 0 ((t − 1) + 1)3 + 1 g (t) = t<1 , t≥2 so g (t) is of the form u1 (t)f (t − 1) for f (t) = (t + 1)3 + 1 = t3 + 3t2 + 3t + 2. Thus L {g (t)} = L {u1 (t)f (t − 1)} = e−s L {f (t)} = e−s L t3 + 3t2 + 3t + 2 = e −s 6 6 3 2 + 3+ 2+ s4 s s s . 4. Find the inverse Laplace transforms of the following functions. (a) 1 s−2 . 1 s−2 L −1 (b) = e2t . 5 + 25 = sin 5t. 5 s2 +25 . L −1 s2 (c) e−2s s2s . +9 L−1 e−2s for f (t) = L−1 s s2 +9 s2 s +9 = u2 (t)f (t − 2), = cos 3t. Thus, L−1 e−2s s s2 + 9 61 = u2 (t) cos (3(t − 2)). (d) 1 s2 −2s+4 . Start by completing the square in the denominator: s2 − 2s + 4 = (s − 1)2 + 3. Thus, L −1 (e) 1 (s − 1)2 + 3 √ 1 3 = √ L−1 (s − 1)2 + 3 3 √ 1 = √ et sin 3t. 3 1 s2 − 2s + 4 = L−1 s+3 s2 +4s+5 . Start by completing the square in the denominator: s2 + 4s + 5 = (s + 2)2 + 1. Thus, L−1 s+3 s2 + 4s + 5 s+3 (s + 2)2 + 1 (s + 2) + 1 (s + 2)2 + 1 s+2 (s + 2)2 + 1 = L−1 = L−1 = L−1 + L−1 1 (s + 2)2 + 1 = e−2t cos t + e−2t sin t. (f) s+3 s2 +3s+3 . Start by completing the square in the denominator: 3 3 s2 + 3s + 3 = (s + )2 + . 2 4 Thus, L−1 s+3 s2 + 3s + 3 = L−1 = L−1 = L−1 s+3 3 (s + 2 )2 + 3 4 (s + 3 ) + 3 2 2 3 (s + 2 )2 + 3 4 3 s+ 2 3 (s + 2 )2 + 3 4 + L−1 (s + 3 2 32 2) + √ 3 4 3 3 √ s+ 2 2 + 3L−1 32 3 32 (s + 2 ) + 4 (s + 2 ) + √ √ √ 3 3 3 3 = e− 2 t cos t + 3e− 2 t sin t. 2 2 = L−1 62 3 4 (g) 12 (s+5)(s−10) . Start by expanding the function by partial fractions: A B (A + B )s + (5B − 10A) 12 = + = . (s + 5)(s − 10) s + 5 s − 10 (s + 5)(s − 10) Equating like terms on both sides, we find A+B =0 5B − 10A = 12 4 This system of linear equations has solution A = − 5 and B = Thus, L−1 12 (s + 5)(s − 10) 41 41 + 5 s + 5 5 s − 10 4 1 4 1 = − L−1 + L−1 5 s+5 5 s − 10 4 4 = − e−5t + e10t . 5 5 = L−1 − 63 4 5. (h) 15 s(s2 −4s+5) . Start by expanding the function by partial fractions: A Bs + C (A + B )s2 + (−4A + C )s + 5A 15 = +2 = . s(s2 − 4s + 5) s s − 4s + 5 s(s2 − 4s + 5) Equating like terms on both sides, we find A+B =0 −4A + C = 0 5A = 15 This system of linear equations has solution A = 3, B = −3 and C = 12. Thus, L−1 s(s2 3 −3s + 12 +2 s s − 4s + 5 3 3s − 12 = L−1 − L −1 2 − 4s + 5 s s 3s − 12 = 3 − L −1 . s2 − 4s + 5 15 − 4s + 5) = L−1 To find the inverse transform of the second function, we first complete the square: s2 − 4s + 5 = (s − 2)2 + 1, so, L−1 3s − 12 − 4s + 5 s2 = L−1 = L−1 = 3L−1 3s − 12 (s − 2)2 + 1 3(s − 2) − 6 (s − 2)2 + 1 s−2 (s − 2)2 + 1 − 6L−1 1 (s − 2)2 + 1 = 3e2t cos t − 6e2t sin 2t. Thus, L−1 s(s2 15 − 4s + 5) 64 = 3 − 3e2t cos t + 6e2t sin 2t. (i) 4s2 +6 s2 (s2 −2s+6) L−1 + 5 s2 −25 . 4s2 + 6 5 +2 2 (s2 − 2s + 6) s s − 25 = L−1 = L−1 4s2 + 6 s2 (s2 − 2s + 6) 4s2 + 6 s2 (s2 − 2s + 6) + L−1 s2 5 − 25 + sinh 5t 2 4s To find L−1 s2 (s2 −+6+6) , we start by expanding the function by 2s partial fractions: 4s2 + 6 Bs + C A1 A2 = + 2+ 2 2 (s2 − 2s + 6) s s s s − 2s + 6 (A1 + B )s3 + (−2A1 + A2 + C )s2 + (6A1 − 2A2 )s + 6A2 = s2 (s2 − 2s + 6) Equating like terms on both sides, we find A1 + B = 0 −2A1 + A2 + C = 4 6A1 − 2A2 = 0 6 A2 = 6 This system of linear equations has solution A1 = 1, A2 = 3, B = −1 and C = 3. Thus, L−1 4s2 + 6 s2 (s2 − 2s + 6) To find L−1 s−3 s2 −2s+6 1 3 −s + 3 + 2+ 2 ss s − 2s + 6 1 1 = L−1 + 3L−1 − L−1 s s2 s−3 = 1 + 3 t − L −1 2 − 2s + 6 s = L−1 s2 s−3 − 2s + 6 , we we first complete the square: s2 − 2s + 6 = (s − 1)2 + 5, so L−1 s2 s−3 − 2s + 6 = L−1 = L−1 s−3 (s − 1)2 + 5 (s − 1) − 2 (s − 1)2 + 5 (s − 1) 2 =L − √ L−1 2+5 (s − 1) 5 √ √ 2 = et cos 5t − √ et sin 5t. 5 −1 65 √ 5 (s − 1)2 + 5 Thus, L−1 4s2 + 6 − 2s + 6) s2 (s2 = 1 + 3t − et cos √ √ 2 5t + √ et sin 5t 5 and L−1 4s2 + 6 5 + − 2s + 6) s2 − 25 s2 (s2 = 1+3t−et cos √ √ 2 5t+ √ et sin 5t+sinh 5t. 5 5. Use convolutions to find the inverse Laplace transform of the following functions. (a) 2 s2 −6s+8 L−1 Let F (s) = forms 2 s2 − 6s + 8 1 s−4 1 s−4 = 2L−1 and G(s) = 1 s−2 . f (t) = L−1 {F (s)} = e4s 1 s−2 These have inverse Laplace transg (t) = L−1 {G(s)} = e2s , so L−1 s2 2 − 6s + 8 = 2L−1 {F (s)G(s)} = 2(f ∗ g )(t) t e4τ e2(t−τ ) dτ =2 0 t e2τ e2t dτ =2 0 t = 2e 2t e2τ dτ 0 = e2t e2τ t 0 = e2t (e2t − 1). 66 . (b) 5 s(s2 +25) . 5 s(s2 + 25) L−1 Let F (s) = forms 1 s 5 s2 +25 . and G(s) = 1 s = L−1 f (t) = L−1 {F (s)} = 1 5 s2 + 25 . These have inverse Laplace transg (t) = L−1 {G(s)} = sin 5t, so L−1 s(s2 5 + 25) = L−1 {F (s)G(s)} = (f ∗ g )(t) t = sin 5τ dτ 0 t 1 = − cos 5τ 5 0 1 1 = − cos 5t + . 5 5 (c) 1 s(s2 +9) . L−1 Let F (s) = forms 1 s 1 s(s2 + 9) and G(s) = = L−1 3 s2 +9 . f (t) = L−1 {F (s)} = 1 1 s 1 s2 + 9 These have inverse Laplace transg (t) = L−1 {G(s)} = sin 3t, so L−1 1 s(s2 + 9) 1 −1 L {F (s)G(s)} 3 1 = (f ∗ g )(t) 3 1t = sin 3τ dτ 30 = t 1 1 − cos 3τ 3 3 0 1 1 = − cos 3t + . 9 9 = 67 . (d) 1 (s−2)2 . L−1 1 (s − 2)2 Let F (s) = G(s) = 1 s−2 . 1 s−2 = L−1 1 s−2 . These have inverse Laplace transforms f (t) = g (t) = e2s , so L−1 1 (s − 2)2 = L−1 {F (s)G(s)} = (f ∗ g )(t) t e2τ e2(t−τ ) dτ = 0 t e2t dτ = 0 t = e2t 1 dτ 0 = te2t . (e) 2 (s−3)2 . L−1 2 (s − 3)2 Let F (s) = G(s) = 1 s−3 . 1 s−3 = 2L−1 1 s−3 These have inverse Laplace transforms f (t) = g (t) = e3t , so L−1 . 2 (s − 3)2 = 2L−1 {F (s)G(s)} = 2(f ∗ g )(t) t e3τ e3(t−τ ) dτ =2 0 t e3t dτ =2 0 t = 2e 3t 1 dτ 0 = 2te3t . 68 6. Use Laplace transforms to solve the following initial value problems. (a) y − 4y = cos t, y (0) = 2. Start by taking the Laplace transform of both sides and solving for Y (s): L {y } − 4L {y } = L {cos t} s sY (s) − 2 − 4Y (s) = 2 s +1 s (s − 4)Y (s) − 2 = 2 s +1 1 1 1 1 Y (s) = +2 = +2 . s − 4 s2 + 1 (s − 4)(s2 + 1) s−4 To find y (t), we take the inverse Laplace transform of Y (s): y (t) = L−1 = L−1 = L−1 We find L−1 tions: 1 1 +2 (s − 4)(s2 + 1) s−4 1 + 2L−1 (s − 4)(s2 + 1) 1 + 2e4t (s − 4)(s2 + 1) 1 (s−4)(s2 +1) 1 s−4 by expanding the function by partial frac- (A + B )s2 + (−4B + C )s + (A − 4C ) 1 A Bs + C = +2 = . 2 + 1) (s − 4)(s s−4 s +1 (s − 4)(s2 + 1) Equating like terms on both sides, we find A+B =0 −4B + C = 0 A − 4C = 1 This system of linear equations has solution A = 4 C = − 17 . Thus, L−1 1 (s − 4)(s2 + 1) 1 −1 L 17 1 4t = e− 17 1 4t = e− 17 = 1 17 , 1 1 s+4 − L−1 s−4 17 s2 + 1 1 −1 s 4 L − L−1 17 s2 + 1 17 1 4 cos t − sin t 17 17 Thus, y (t) = 1 4t e − cos t − 4 sin t + 2e4t . 17 69 1 B = − 17 and 1 s2 + 1 (b) y − 6y = e3t , y (1) = 0. Here, we are not given y (0), so we will set it to a constant, c0 . After finding the solution, we can then use the given initial condition to solve for c0 . Take Laplace transforms of both sides and solve for Y (s): 1 s−3 1 c0 = + . (s − 6)(s − 3) s − 6 sY (s) − c0 − 6Y (s) = Y (s) = 1 1 + c0 s−6 s−3 Now, taking the inverse Laplace transform yields 1 (s − 6)(s − 3) y (t) = L−1 We find L−1 1 (s−6)(s−3) + c0 e6t . using convolutions. Let F (s) = 1 s−6 1 s−3 . and G(s) = These have inverse transforms f (t) = e6t and g (t) = e3t respectively. Then L−1 1 (s − 6)(s − 3) = L−1 {F (s)G(s)} = (f ∗ g )(t) t e6τ e3(t−τ ) dτ = 0 t = e3t e3τ dτ 0 = e3t 1 3τ e 3 t 0 1 = e3t (e3t − 1). 3 Thus, 1 6t 1 3t e − e + c0 e6t . 3 3 Lastly, we use y (1) = 0 to solve for c0 : y (t) = 16 13 e − e + c0 e6 3 3 1 1 −3 c0 = − e , 33 0 = y (1) = so y (t) = 1 6t 1 3t 11 1 1 e − e + ( − e−3 )e6t = − e3t + e6t−3 . 3 3 33 3 3 70 (c) y + y = sin t, y (2π ) = 2. Here, we are not given y (0), so we will set it to a constant, c0 . After finding the solution, we can then use the given initial condition to solve for c0 . Take Laplace transforms of both sides and solve for Y (s): 1 s2 + 1 1 c0 = + . 2 + 1) (s + 1)(s s+1 sY (s) − c0 + Y (s) = Y (s) = 1 1 + c0 2+1 s+1 s Now, taking the inverse Laplace transform yields y (t) = L−1 We find L−1 1 (s+1)(s2 +1) 1 (s + 1)(s2 + 1) + c 0 e −t . by expanding in partial fractions: 1 A Bs + C (A + B )s2 + (B + C )s + (A + C ) = +2 = . (s + 1)(s2 + 1) s+1 s +1 (s − 4)(s2 + 1) Equating like terms on both sides, we find A+B =0 B+C =0 A+C =1 This system of linear equations has solution A = C = 1 . Thus, 2 L−1 1 (s + 1)(s2 + 1) 1 −1 1 1 L − L−1 2 s+1 2 1 −t 1 1 = e − cos t + sin t 2 2 2 = 1 2, s s2 + 1 B = − 1 and 2 + Thus, 1 −t 1 1 e − cos t + sin t + c0 e−t . 2 2 2 Lastly, we use y (2π ) = 2 to solve for c0 : y (t) = 1 −2π 1 e − + c0 e−2π 2 2 5 2π 1 c0 = e − , 2 2 2 = y (2π ) = so 1 −t 1 1 5 1 e − cos t + sin t + ( e2π − )e−t 2 2 2 2 2 1 1 5 = − cos t + sin t + e2π−t . 2 2 2 y (t) = 71 1 −1 L 2 1 s2 + 1 1 0≤t<π , y (0) = y (0) = 0. 0 t≥π Start by rewriting the right hand side in terms of the unit step function: y + 2y + 5y = 1 − uπ (t). (d) y + 2y + 5y = Now, taking Laplace transforms of both sides gives: (s2 Y (s) − sy (0) − y (0)) + 2(sY (s) − y (0)) + 5Y (s) = s2 Y (s) + 2sY (s) + 5Y (s) = Y (s) = 1 e−πs − s s 1 e−πs − s s 1 e−πs 1 e−πs 1 − = − . s2 + 2s + 5 s s s(s2 + 2s + 5) s(s2 + 2s + 5) We proceed by finding the inverse Laplace transforms of the functions 1 on the right. Start with the first: s(s2 +2s+5) . We expand this using partial fractions: A Bs + C (A + B )s2 + (2A + C )s + 5A 1 = +2 = . s(s2 + 2s + 5) s s + 2s + 5 s(s2 + 2s + 5) Equating like terms on both sides, we find A+B =0 2A + C = 0 5A = 1 This system of linear equations has solution A = C = − 2 . Thus, 5 1 s(s2 + 2s + 5) = 1 −1 L 5 = L−1 1 s 1 1 − L−1 5 5 − 1 −1 L 5 1 5, B 1 = − 5 , and s+2 s2 + 2s + 5 (s + 1) + 1 (s + 1)2 + 4 (s + 1) 1 1 1 −1 − L−1 − L 5 5 (s + 1)2 + 4 10 1 1 1 −t = − e−t cos 2t − e sin 2t 5 5 10 = To find the inverse transform of L −1 s(s2 e−πs + 2s + 5) 72 e−πs s(s2 +2s+5) , = L−1 e−πs 2 (s + 1)2 + 4 we observe s(s2 1 + 2s + 5) . We just found that L−1 so L−1 e−πs 1 s(s2 + 2s + 5) 1 s(s2 +2s+5) = 1 5 1 − 1 e−t cos 2t − 10 e−t sin 2t, 5 1 1 −(t−π) 1 − e−(t−π) cos (2(t − π )) − e sin (2(t − π )) 5 5 10 1 1 1 −t+π = + e−t+π cos 2t + e sin 2t. 5 5 10 = Thus, 1 1 −t 1 11 1 − e cos 2t − e−t sin 2t − − e−t+π cos 2t − e−t+π sin 2t 55 10 55 10 1 −t 1 −t π π = − e (1 + e ) cos 2t − e (1 + e ) sin 2t 5 10 y (t) = 73 t 5 0≤t<5 , y (0) = y (0) = 0. 0 1≥5 Start by rewriting the right hand side in terms of the unit step function: t t y + 2y + 5y = + 1 − u5 (t). 5 5 (e) y + 2y + 5y = Now, taking Laplace transforms of both sides gives: 1 1 −5s 11 − e 2 5s 5 s2 1 1 1 1 = − e−5s 2 (s2 + 1) 2 (s2 + 1) 5s 5s s2 Y (s) + Y (s) = Y (s) = s2 1 1 1 −5s 11 − e 2 +1 5s 5 s2 We need to find the inverse Laplace transform of both sides. Start with s2 (s1 +1) . We find this using convolutions: let F (s) = s1 and 2 2 G(s) = s21 . These have inverses f (t) = t and g (t) = sin(t) respec+1 tively. Then 1 L−1 s2 (s2 + 1) = L−1 {F (s)G(s)} = (f ∗ g )(t) t (t − τ ) sin τ dτ = 0 t t sin τ dτ − =t 0 τ sin τ dτ 0 t t t = t [− cos τ ]0 + [τ cos τ ]0 − cos τ dτ 0 t = t(1 − cos t) + t cos t − [sin τ ]0 = t − sin t Next, we observe that L−1 1 e−5s s2 (s2 + 1) so y (t) = = u5 (t) ((t − 5) − sin (t − 5)) , 1 [t − sin t − u5 (t) (t − 5 − sin (t − 5))] . 5 74 (f) y + 4y = sin 3t + 8δ π (t), y (0) = 3, y (0) = 0. 2 We begin by taking the Laplace transform of both sides and solving for Y (s): s2 Y (s) − 3s + 4Y (s) = −π 3 + 8e 2 s s2 + 9 −π 3 1 + 8 e 2 s + 3s 2+9 +4 s −π s 3 1 + 8e 2 s 2 +3 2 . =2 (s + 4)(s2 + 9) s +4 s +4 Y (s) = s2 We proceed by finding the inverse Laplace transform of both sides: y (t) = L−1 3 (s2 + 4)(s2 + 9) + 8u π (t) sin 2 t − 2 We find the inverse transform of 3 (s2 +4)(s2 +9) π 2 + 3 cos 2t. using partial fractions: 3 As + B Cs + D =2 +2 (s2 + 4)(s2 + 9) s +9 s +4 = (A + C )s3 + (B + D)s2 + (4A + 9C )s + (4B + 9D) . (s2 + 9)(s2 + 4) Equating like terms on both sides, we find A+C =0 B+D =0 4A + 9C = 0 4B + 9 D = 3 3 This system of linear equations has solution A = C = 0, B = − 5 , 3 and D = 5 . Thus, L−1 3 (s2 + 4)(s2 + 9) 3 1 3 = − L−1 + L−1 5 s2 + 9 5 3 3 = − sin 3t + sin 2t. 5 5 1 s2 + 4 Thus, 3 π 3 y (t) = − sin 3t + sin 2t + 8u π (t) sin 2 t − 2 5 5 2 75 + 3 cos 2t. (g) y − y = g (t) + δ (t − 3)et , y (0) = y (0) = 0, where g (t) is defined by g (t) = t 2 0≤t<1 . t≥1 Start by writing g (t) as in terms of the unit step function: g (t) = t + (2 − t)u1 (t) = t + (1 − (t − 1))u1 (t). Then, taking Laplace transforms of both sides and solving for Y (s), we have 1 1 1 s2 Y (s) − sY (s) = 2 + e−s − 2 + e−3(t−1) s ss 1 1 1 1 Y (s) = + e−s − + e−3(s−1) s(s − 1) s2 s s2 1 1 1 1 =3 + e −s 2 − e−s 3 + e3 e−3s s (s − 1) s (s − 1) s (s − 1) s(s − 1) To find these Laplace transforms, we observe that 1 s3 (s−1) can be expanded in partial fractions as: 1 1 s(s−1) , s2 (s−1) and 1 1 1 = − s(s − 1) s−1 s 1 1 1 1 = −− s2 (s − 1) s − 1 s s2 1 1 1 1 1 = − − 2− 3 3 (s − 1) s s−1 s s s These have inverse Laplace transforms 1 = et − 1 s(s − 1) 1 = et − 1 − t L−1 s2 (s − 1) 1 t2 L−1 = et − 1 − t − . s3 (s − 1) 2 L −1 Therefore, L−1 e−3s L−1 e−s L−1 1 s(s − 1) = u3 (t) et−3 − 1 1 = u1 (t) et−1 − 1 − (t − 1) = u1 (t) et−1 − t) − 1) 1 (t − 1)2 e−s 3 = u1 (t) et−1 − 1 − (t − 1) − s (s − 1) 2 2 t +1 = u1 (t) et−1 − . 2 s2 (s 76 Finally, t2 + u1 (t) et−1 − t) 2 t2 + 1 − u1 (t) et−1 − 2 y (t) = et − 1 − t − 77 + e3 u3 (t) et−3 − 1 . 0.4 Linear Algebra 1. Solve the the system of equations x2 + x3 + x4 = 2 x1 + x2 − x4 = 0 2x1 − x3 − 4x4 = −2 x1 + 2x2 + 2x3 = 4 −x1 + x2 + 3x4 = 0 The augmented matrix of this system is 0 1 2 1 −1 1 1 0 2 1 1 0 −1 2 0 1 −1 −4 0 3 2 0 −2 4 0 . We proceed by row reducing the matrix: 0 1 2 1 −1 −→ 1 1 0 2 1 1 0 0 0 0 1 0 −1 2 0 1 1 −2 1 2 1 −1 −4 0 3 0 1 −1 2 0 2 0 −2 4 0 −1 1 −2 1 2 −→ 0 2 −2 4 0 1 0 −→ 0 0 0 0 1 0 0 0 1 0 2 1 −1 −→ 0 0 1 0 0 −2 1 0 0 0 1 1 0 2 1 1 0 0 0 0 0 0 2 0 0 0 1 −1 2 0 0 −1 1 1 0 1 0 1 0 −2 −1 1 −4 0 3 −2 1 0 0 0 0 2 −2 4 0 −2 2 2 2 −4 Translating this back into equations yields: x1 − 2 x4 = 0 x2 + x4 = 0 x3 = 2 with x4 free. Set x4 = t ∈ R. Then, x1 = 2x4 = 2t and x2 = −x4 = −t. Thus, the solution to the system is 78 x1 x2 x3 x4 = 2t −t 2 t . 10 1 1 2. Let A = 1 2 02 −2 1 4 6 2 2 0 −2 4 5 2 −2 1 −1 . −1 −2 (a) Find a basis for the row space of A. What is rank(A)? (b) Find a basis for the column space of A. (c) Find a basis for the kernel of A. To answer all of these questions, you must first use the Gaussian algorithm to put the matrix in reduced row echelon form. This gives 1 0 −2 2 4 1 1 0 −2 0 0 3 1 1 1 2 5 −1 −→ · · · −→ 0 1 3 0 1 −2 . A= 1 2 4 0 0 0 1 2 −1 0 2 −1 0 2 6 −2 −2 −2 00 0 00 0 (a) The non-zero rows of the reduced row echelon matrix (those with leading 1’s) give a basis for the row space: {1 0 −2 0 0 3,0 1 3 0 1 −2 , 0 0 0 1 2 −1 } . The rank of A is equal to the dimension of the row space, so rank(A)=3. (b) The find a basis for the column space, we look to see what columns in the reduced row echelon matrix contain leading 1’s. The corresponding columns in the original matrix form a basis for the column space. Here, there are leading 1’s in the first, second and fourth columns of the row reduced matrix, so a basis for the column space is: 2 0 1 1 1 2 , , . 0 1 2 −2 2 0 79 (c) To find a basis for the kernel of A, we must solve AX = 0. From the reduced matrix, this is equivalent to solving the system of linear equations: x1 − 2x3 + 3x6 = 0 x2 + 3x3 + x5 − 2x6 = 0 x4 + 2x5 − x6 = 0, where x3 = r, x5 = s and x6 = t are free parameters. Thus, x1 = 2x3 − 3x6 = 2r − 3t x2 = −3x3 − x5 + 2x6 = −3r − s + 2t x4 = −2x5 + x6 = −s + t Thus, X = 2r − 3t −3r − s + 2t sis for ker(A) is: {2 −3 1 −2 0 0 T ,0 80 −1 r −2r + t 0 0 1 0 s T t T , −3 , and a ba- 2 0 1 0 1 T }. 1 1 3. Let D = 0 −1 11 11 00 1 −3 0 1 . 1 0 (a) Find the reduced row echelon form for D, and determine the rank of D and the dimension of the kernel of A. (b) Find a basis for the row space of D. (c) Find a basis for the column space of D. (d) Find a basis for the kernel of D. (a) The reduced row echelon form of 10 0 1 0 0 00 D is 2 −1 0 0 0 0 . 1 0 Since there are three leading 1’s in the reduced matrix, rank(D) = 3. By the rank theorem, rank(D) + dim(ker(D)) = 4, so the kernel has dimension 1. (b) The non-zero rows of the reduced matrix found in part (a) give a basis for the row space: {1 0 2 0,0 1 −1 0,0 0 0 1 }. (c) The find a basis for the column space, we look to see what columns in the reduced row echelon matrix contain leading 1’s. The corresponding columns in the original matrix form a basis for the column space. Thus, a basis is: 1 0 1 1 1 1 , , . 1 0 0 −1 1 0 81 (d) To find a basis for the kernel, we must solve DX = 0. From the reduced matrix, this is equivalent to solving the system of linear equations: x1 + 2x3 = 0 x2 − x3 = 0 x4 = 0, where x3 = t is free. We thus have x1 = −2t and x2 = t, so −2 −2t x1 1 x2 t X= = x3 t = t 1 , 0 0 x4 −2 1 1 0 so a basis for the kernel is 82 . 5 4. Let A = 1 0 0 0 0 0 3 . −2 (a) Find all eigenvalues of A. (b) Find the corresponding eigenvectors. (c) Find a diagonal matrix D and an invertible matrix P such that P −1 AP = D. (a) Start by finding the characteristic polynomial: det(A − λI ) = 5−λ 1 0 0 −λ 0 0 3 = λ(5 − λ)(λ + 2). −2 − λ Thus det(A − λI ) = 0 if and only if λ = 0, 5, −2. These are the three eigenvalues. (b) To find the eigenvectors, we must find all solutions to (A − λ)X = 0 for each of the eigenvalues. • For λ1 = 0: 5 A − 0I = 1 0 0 0 0 0 3 , −2 so (A − 0I )X = 0 if 5x1 = 0 x1 + 3 x3 = 0 −2x3 = 0 Thus, x1 = x3 = 0 and x2 = t is free. The eigenspace for λ1 = 0 is thus 0 0 t = t 1 . 0 0 for t ∈ R. 83 • For λ2 = 5: 0 A − 5I = 1 0 0 −5 0 0 3 , −7 so (A − 5I )X = 0 if 0x1 = 0 x1 − 5x2 + 3x3 = 0 −7x3 = 0 t Thus, x1 = t is free, x2 = 5 , and x3 = 0. The eigenspace for λ2 = 5 is thus t 5 t t = 1. 5 5 0 0 for t ∈ R. • For λ3 = −2: 7 A + 2I = 1 0 0 2 0 0 3 , 0 so (A + 2I )X = 0 if 7 x1 = 0 x1 + 2 x2 + 3 x3 = 0 0 x3 = 0 Thus, x3 = t is free, x1 = 0 and x2 = − 3t . The eigenspace for 2 λ3 = −2 is thus 0 0 − 3t = t −3 . 2 2 2 t for t ∈ R. (c) The matrix has three distinct eigenvalues and is thus invertible. The matrix D is a diagonal matrix with the three eigenvalues on the main diagonal, and P is a matrix which has the corresponding eigenvectors as its columns: 00 0 05 0 D = 0 5 0 and P = 1 1 −3 . 0 0 −1 00 2 84 3 5. Let A = t 0 −1 3 0 0 0 . 2 (a) Find all values of t for which A is diagonalizable. (b) In the case where t = −1, find a basis of eigenvalues for R3 . (c) For t = −1, find a diagonal matrix D and an invertible matrix P such that P −1 AP = D. (a) Start by finding the characteristic polynomial: det(A − λI ) = 3−λ t 0 −1 0 3−λ 0 = (2 − λ)((3 − λ)2 + t). 0 2−λ Thus, λ1 = 2 is √ always an eigenvalue of A. (3 − λ)2 + t = 0 if and only if λ = 3 ± −t. Thus, if t = 0, 3 is a repeated root, while if t = −1, then 2 is a repeated root. Thus, for t = 0, −1, A has three distinct eigenvalues and is diagonalizable. For t = 0, −1, we must find the eigenvectors. We do t = 0 here; t = −1 is done in part (b). 3 −1 0 If t = 0, A = 0 3 0 , and λ = 3 is an eigenvalue of multiplicity 002 2. A is only diagonalizable if there are two linearlly independent eigenvectors for λ = 3. We must therefore solve (A − 3I )X = 0: 0 −1 0 0 , A − 4I = 0 0 0 0 −1 so (A − 3I )X = 0 if −x 2 = 0 0=0 −x 3 = 0 Thus, x1 = t is free and x2 = x3 = 0. The eigenspace for λ = 3 is thus t 1 0 = t 0 . 0 0 for t ∈ R. Since there is only one linearly independent vector in the eigenspace, the matrix is not diagonalizable. 85 3 −1 0 (b) When t = −1, A = −1 3 0 , and λ = 2 is an eigenvalue of 0 02 multiplicity 2. To find the eigenvectors, we must find all solutions to (A − λ)X = 0 for each of the eigenvalues λ1 = λ2 = 2, λ3 = 4: • For λ1 = λ2 = 2: 1 A − 2I = −1 0 −1 1 0 0 0 , 0 so (A − 2I )X = 0 if x1 − x2 = 0 −x 1 + x 2 = 0 0x3 = 0 Thus, x1 = s and x3 = t are free and x2 = s. The eigenspace for λ = 2 is thus s 1 0 s = s 1 + t 0 . t 0 1 for s, t ∈ R. Because there are two linearly independent eigenvectors, we see that A is diagonalizable when t = −1. • For λ3 = 4: −1 −1 0 A − 4I = −1 −1 0 , 0 0 −2 so (A − 4I )X = 0 if −x 1 − x 2 = 0 −x 1 − x 2 = 0 −2x3 = 0 Thus, x1 = t is free, x2 = −t, and x3 = 0. The eigenspace for λ3 = 4 is thus t 1 −t = t −1 . 0 0 for t ∈ R. Thus, 0 1 1 1 , 0 , −1 0 1 0 is a basis of eigenvectors for R3 . 86 (c) The matrix has three distinct eigenvalues and is thus invertible. The matrix D is a diagonal matrix with the three eigenvalues on the main diagonal, and P is a matrix which has the corresponding eigenvectors as its columns: 200 10 1 D = 0 2 0 and P = 1 0 −1 . 004 01 0 6. Use the Gram-Schmidt algorithm to construct an orthonormal basis for span({X1 , X2 , X3 }), where X1 = 1 1 0 1 , X2 = 0 1 1 1 −1 1 1 1 , X3 = . We apply the Gram-Schmidt Algorithm: F1 = X1 = F2 = X2 − = 0 1 1 1 −1 1 1 1 so F1 = √ 3 X2 F1 F1 F1 2 2 − 3 F3 = X3 − = 1 1 0 1 1 1 0 1 = −2 3 1 3 1 so F1 = 1 3 X3 F2 X3 F1 F1 − F2 2 F1 F2 2 2 −2 − 5 1 11 7 13 5 1− 3 = −2 − 1 5 30 5 1 1 1 3 Hence, an orthonormal basis is 1 F2 F3 F1 , , =√ 3 F1 F2 F3 87 5 3 so F1 = 2 5 5 1 1 0 1 , 2 − 3 13 3 , 1 5 1 3 − 2 5 1 5 5 . 2 2 −15 5 1 7. Let A = −1 0 −1 2 −1 0 −1. Find: 1 (a) the characteristic polynomial of A. (b) the eigenvalues of A. (c) the corresponding eigenvalues. (d) an orthogonal matrix U and a diagonal matrix D so that U T AU = D. (a) det(A − λI ) = 1−λ −1 0 −1 2−λ −1 0 −1 = −λ(λ − 3)(λ − 1). λ (b) The eigenvalues of A are the roots of the characteristic polynomial, which are λ1 = 0, λ2 = 1, and λ3 = 3. (c) To find the corresponding eigenvectors, we solve (A − λI ) = 0 for each of the eigenvalues. • For λ1 = 0, 1 A − 0I = −1 0 1 0 −1 −→ · · · −→ 0 0 1 −1 2 −1 0 −1 0 −1 1 0 so x1 = x2 = x3 = t, t ∈ R: x1 x2 x3 t t t = Hence, E0 (A) has basis X1 = 1 1 1 . • For λ2 = 1, 0 1 −1 −→ · · · −→ 0 0 0 −1 1 −1 0 A − 1I = −1 0 so x1 + x3 = 0, x2 = 0, x3 = t ∈ R: x1 x2 x3 = −t 0 t Hence, E1 (A) has basis X2 = 88 1 0 −1 . 0 1 0 1 0 0 • For λ3 = 3, −2 A − 3I = −1 0 0 1 −1 −→ · · · −→ 0 −2 0 −1 −1 −1 −1 2 0 0 1 0 so x1 − x3 = 0, x2 + 2x3 = 0, x3 = t ∈ R: x1 x2 x3 = t −2 t Hence, E3 (A) has basis 1 −2 1 X3 = . (d) D will be a matrix with the eigenvalues along the main diagonal. To find U , we must find an orthonormal basis for R3 consisting of eigenvectors. These vectors are then the columns of U . Now, we must use {X1 , X2 , X3 } to obtain an orthogonal basis {F1 , F2 , F3 } for R3 . Since each eigenvector corresponds to a distinct eigenvector, we can conclude they are orthogonal vectors. Hence, dividing each vector by it’s norm yields an orthogonal basis: Y1 = 1 √ 3 1 √ 3 1 √ 3 , Y2 = 1 √ 2 0 1 − √2 , Y3 = 1 √ 6 2 − √6 1 √ 6 . Thus: 0 D = 0 0 0 1 0 0 0 3 1 √ √3 1 3 1 √ 3 and 89 U= 1 √ 2 0 1 − √2 1 √ 6 2 − √6 . 1 √ 6 10 2 8. Let A = 0 2 0 . Find an orthogonal matrix Q and a diagonal matrix 2 0 −2 D so that QDQT = A. We start by finding the characteristic polynomial of A: det(A − λI ) = 1−λ 0 −2 0 2−λ 0 2 0 = −(λ + 3)(λ − 2)2 . λ+2 Thus, the matrix has eigenvalues λ1 = −3 and λ2 = λ3 = 2. To find the corresponding eigenvectors, we solve (A − λI ) = 0 for each of the eigenvalues. • For λ1 = −3, 4 A + 3I = 0 2 2 0 . 1 0 5 0 so 4x1 − 2x3 = 0, 5x2 = 0 and 2x1 + x3 = 0. Thus x1 x2 x3 = −t 0 2t t ∈ R. , Hence, E−3 (A) has basis X1 = −1 0 2 . • For λ2 = λ3 = 2, 02 0 0 04 −1 A − 2I = 0 −2 so −x1 + 2x3 = 0, 2x1 − 4x3 = 0 and x2 = t ∈ R is free, so the solutions are x1 2s x2 = t x 3 2 Hence, E2 (A) has basis X2 = 2 0 1 X3 = 0 1 0 . Now, we must use {X1 , X2 , X3 } to obtain an orthogonal basis {F1 , F2 , F3 } for R3 . Since X1 corresponds to a different eigenvector from X2 and X3 , we can conclude it is orthogonal to both vectors. We also observe that X2 X3 = 0, so the vectors are already orthogonal (otherwise, we would apply the Gram-Schmidt algorithm to the set {X2 , X3 }.) 90 To find an orthonormal basis for R3 , we divide each vector Xi by its norm to obtain {Y1 = 1 − √5 0 2 √ 5 , Y2 = 2 √ 5 0 1 √ 5 , Y3 = The columns of Q are the vectors Y1 , Y2 , Y3 : √ − 15 Q= 0 1 √ 5 while −3 D= 0 0 91 2 √ 5 0 1 √ 5 0 1 . 0 00 2 0 . 02 0 1 0 }. 9. Consider the matrix A = 1 −2 1 . 4 (a) Diagonalize A. (b) Evaluate eAt . (c) Using eAt , write down an expression for the solution to the system of linear equations: y1 = y1 + y2 y2 = −2y1 + 4y2 Subject to the initial condition y1 (0) = a and y2 (0) = b. (a) We start by finding the eigenvalues of A, using the characteristic polynomial: det(A − λI ) = 1−λ −2 1 4−λ = λ2 − 5λ + 6 = (λ − 2)(λ − 3) Thus, the eigenvalues of A are the roots of the characteristic polynomial, λ1 = 2 and λ2 = 3. To find the corresponding eigenvectors, we solve (A − λI ) = 0 for each of the eigenvalues. • For λ1 = 2, A − 2I = −1 −2 1 −1 −→ 2 0 1 . 0 Thus, x1 = x2 = t, for any t ∈ R. Thus, a basis for the eigenspace for λ1 is [ 1 ]. 1 • For λ2 = 3, A − 3I = −2 −2 1 −2 −→ 1 0 1 . 0 Thus, x1 = t and x2 = 2t, for any t ∈ R. Thus, a basis for the eigenspace for λ1 is [ 1 ]. 2 Thus, taking P = 1 1 1 2 and D = 2 0 D. 92 0 , we have that P −1 AP = 3 (b) eAt = P e2t 0 1 2 0 P −1 e3t e2t 0 = 1 1 = 2e2t − e3t 2e2t − 2e3t 0 e3t 2 −1 −1 1 −e2t + e3t −e2t + 2e3t (c) The system has solution y(t) = eAt y(0) = 2e2t − e3t 2e2t − 2e3t = (2a − b)e2t + (b − a)e3t (2a − b)e2t + (2b − 2a)e3t 93 −e2t + e3t −e2t + 2e3t a b 10. Solve the system of linear differential equations y1 = y1 + 3 y2 y2 = 4y1 + 2y2 subject to the initial conditions y1 (0) = 1 and y2 (0) = 6. We start by finding the eigenvalues of A, using the characteristic polynomial: det(A − λI ) = 1−λ −2 1 4−λ = λ2 − 3λ − 10 = (λ − 5)(λ + 2) Thus, the eigenvalues of A are the roots of the characteristic polynomial, λ1 = 5 and λ2 = −2. To find the corresponding eigenvectors, we solve (A − λI ) = 0 for each of the eigenvalues. • For λ1 = 5, 4 −4 A − 2I = −3 4 −→ 3 0 3 . 0 Thus, if x2 = 4t, for t ∈ R, then x1 = 3t. Thus, a basis for the eigenspace for λ1 is [ 3 ]. 4 • For λ2 = −2, A − 3I = −3 −4 −3 −3 −→ −4 0 −3 . 0 Thus, x1 = t and x2 = −t, for any t ∈ R. Thus, a basis for the eigenspace for λ1 is −1 . 1 3 Thus, taking P = 4 so −1 5 and D = 1 0 eAt = P 1 7 1 = 7 = e5t 0 3 4 0 , we have that P −1 AP = D, −2 0 P −1 e−2t −1 1 e5t 0 3e5t + 4e−2t 4e5t − 4e−2t 94 0 e−2t 1 −4 3e5t − e−2t 4e2t + 3e−2t 1 3 Thus, the system has solution y(t) = eAt y(0) 1 7 1 = 7 = = 3e5t + 4e−2t 4e5t − 4e−2t 21e5t − 14e−2t 28e2x + 14e3x 3e5t − 2e−2t 4e2x + 2e3x 95 3e5t − e−2t 4e2t + 3e−2t 1 6 ...
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This note was uploaded on 12/01/2010 for the course MATH 263 taught by Professor Sidneytrudeau during the Fall '09 term at McGill.

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