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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 + 33 + ln v − 1 = ln kx−4
ln v + 33 v − 1 = ln kx−4
v + 33 v − 1 = kx−4
y
y
 + 33  − 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 kx3
v= 3 ln kx3 y = x 3 ln kx3 . 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 ﬁrstorder linear diﬀerential equations.
(a) dy
dx y
+ x − x5 = 0
This is ﬁrstorder 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 nonhomogeneous 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 CauchyEuler 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 nonhomogeneous CauchyEuler
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 CauchyEuler 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 CauchyEuler 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 CauchyEuler 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 CauchyEuler 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 CauchyEuler 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 nonzero 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 nonzero 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 nonzero 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 nonzero 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 GramSchmidt 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 GramSchmidt 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 GramSchmidt 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.
 Fall '09
 SidneyTrudeau
 Math, Equations

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