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### SP07Finals-3

Course: M 427K, Spring 2007
School: University of Texas
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- M427K Final May 7, 2007 Linear Equations Method of Integrating Factors y + p(t)y = q(t) (y) (t)q(t)dt + C (t) (t) = e p(t)dt = (t)q(t) y(t) = Separable Equations When the ODE can be framed in the following form: y dy g(y) dy g(y) = f (t)g(y) = f (t)dt = f (t)dt Check if the funstion you are dividing with = 0 or not and check to see if such a function satisfies the original equation. Homogeneous...

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- M427K Final May 7, 2007 Linear Equations Method of Integrating Factors y + p(t)y = q(t) (y) (t)q(t)dt + C (t) (t) = e p(t)dt = (t)q(t) y(t) = Separable Equations When the ODE can be framed in the following form: y dy g(y) dy g(y) = f (t)g(y) = f (t)dt = f (t)dt Check if the funstion you are dividing with = 0 or not and check to see if such a function satisfies the original equation. Homogeneous Equations When the ODE can be put in the following form y y = = f (y/x) xv + v v = y/x y = vx xv + v xv dv f (v) - v = = = f (v) f (v) - v dx x Check if solution to f (v) = v satifsies the original ODE. 1 Non-Linear Equations Bernoulli Equations First Order ODEs M (x, y)dx + N (x, y)dy = 0. Need to find (x, y) such that x (x, y) = M (x, y) and y (x, y) = N (x, y). Then (x, y) = C gives the implicit for, of the solution. Exact Equations If My = Nx , the ODE is EXACT. Then (x, y) = M (x, y)dx + h(y) and y (x, y) = N (x, y) = ( M (x, y)dx)y + h (y). Then solve for h(y) to get the M (x, y)dx + h(y) = C. OR If My = Nx , the ODE is EXACT. Then (x, y) = N (x, y)dy + g(x) and x (x, y) = M (x, y) = ( N (x, y)dy)x + g(x). Then solve for g(x) to get the N (x, y)dy + g(x) = C. OR If My = Nx , the ODE is EXACT. Then (x, y) = M (x, y)dx + h(y) and (x, y) = N (x, y)dy+g(x). Then solve for h(y) and g(x) to get the M (x, y)dx+ h(y) = C OR N (x, y)dy + g(x) = C. Either of these expressions work. Non-Exact Equations - Integrating Factors If My = Nx , the ODE is NOT EXACT. Then need to find an integrating factor, which when multiplied to the ODE turns the ODE EXACT. If My -Nx N is a function of x, then the integrating factor is (x) = exp( My -Nx dx) N OR If My -Nx -M is a function of y, then the integrating factor is (y) = exp( My -Nx -M dy) Multiply the original equation with the integrating factor thus turning it into an EXACT ODE and then solve it using the known method of solving EXACT ODEs. Second Order Linear ODEs Wronskian and Linear Independence: If y1 , y2 are two solutions to the 2nd order ODE, then W (y1 , y2 ) = y1 y2 -y1 y2 . If the given ODE is of the form y +p(t)y +q(t)y = g(t), then the Wronskian is given by W (y1 , y2 ) = Cexp(- p(t)dt) Homogeneous Equations w/ Constant Coefficients 2 Characteristic Equation in r with real distinct roots r1 , r2 General Solution y = C1 er1 t + C2 er2 t Characteristic Equation in r with repeated roots r - Reduction of Order General Solution y = C1 ert + C2 tert Characteristic Equation in r with complex roots r = General Solution y = et( C1 cos(t) + C2 sin(t)) Reduction of Order If one of the solutions is known y1 then. let y2 = vy1 . Make this substitution into the original ODE to get a separable ODE in v. Solve this ODE for v to get the other solution y2 = vy1 . Inhomogeneous Equations w/ Constant Coefficients Method of Undetermined Coefficients PAGE: 164 (7) y -2y +2y = 0 characteristic equation is r 2 -2r+2 = 0 (r-1)1 +1 = 0 r = 1i. So, the general solution is y(t) = c1 et cos(t) + c2 et sin(t). PAGE: 172 (1) y -2y +y = 0 characteristic equation is r 2 -2r+1 = 0 r = 1 y(t) = c1 et +c2 tet . (4) 4y + 12y + 9y = 0 characteristic equation is 4r 2 + 12r + 9 = 0 (2r + 3)2 = 0 r = -3/2 y(t) = c1 e( - 3t/2) + c2 te( - 3t/2). (5) y - 2y + 10y = 0 characteristic equation is r 2 - 2r + 10 = 0 (r - 1)2 + 9 = 0 r = 1 3i y(t) = c1 et cos(3t) + c2 et sin(3t). PAGE: 184 (1) y - 2y - 3y = 3e2t Homogeneous Solution: y -2y -3y = 0 characteristic equation is r 2 -2r-3 = 0 r = 3, -1 yh = c1 e3t +c2 e-t 3 Particular Solution (Method of Undetermined Coefficients): yp = y = Ae2t y = 2Ae2t , y = 4Ae2t (4A - 2(2A) - 3A)e2t = 3e2t -3A = 3 A = -1. So, yp = -e2t y(t) = yh + yp = c1 e3t + c2 e-t - e2t is the general solution. (3) y - 2y - 3y = -3te-t Homogeneous Solution: y -2y -3y = 0 characteristic equation is r 2 -2r-3 = 0 r = 3, -1 yh = c1 e3t +c2 e-t Particular Solution (Method of Undetermined Coefficients - e-t is already a homogeneous solution): yp = y = (At2 + Bt)e-t y = (2At + B - At2 - Bt)e-t = (-At2 + t(2A - B) + B)e-t y = (-2At + 2A - B + At2 - t(2A - B) - B)e-t = (At2 - t(4A - B) - 2B + 2A)e-t Plugging them into the original equation: (At2 - t(4A - B) - 2B + 2A - 2(-At2 + t(2A - B) + B) - 3(At2 + Bt))e-t = -3te-t 3 A = 2B; A = 3/8 B = 3/16. So, yp = ( 3 t2 + 16 t)e-t . 8 y = yh + yp = c1 e3t + c2 e-t + ( 3 t2 + 8 (4) y + 2y = 3 + 4sin(2t) Homegeneous Solution: y + 2y = 0 r 2 + 2r = 0 r = 0, -2 yh = c1 + c2 e-2t Particular Solution (Split into 2 particular solutions: 1 is already a homogeneous solution): yp1 = At2 + Bt and yp2 = Dsin(2t) + Ecos(2t). So: 4A + 2(2At + B) = 3 A = 0andB = 3 2 3 -t 16 t)e is the general solution. -4(Dsin(2t) + Ecos(2t)) + 2(2Dcos(2t) - 2Esin(2t)) = 4sin(2t) D + E = -1; D = E D = E = -1 2 4 So, putting all the particular solutions and the homogeneous solutions together: 3 y = yh + yp1 + yp2 = c1 + c2 e-2t + 2 t - 1 (sin(2t) + cos(2t)) 2 (6) y + 2y + y = 2e-t Homogeneous Solution: y + 2y + y = 0 r 2 + 2r + 1 = 0 r = -1 yh = c1 e-t + c2 te-t Particular Solution: yp = y = At2 e-t A = 1 (after computing y , y and plugging them back into the original equation). y = yh + yp = c1 e-t + c2 te-t + t2 e-t is the general solution. PAGE: 190 (7) y + 4y + 4y = t-2 e-2t = g(t) We cannot use the method of undetermined coefficients as the right hand function CANNOT be written in the form of undetermined coefficients. So, need to use the method of variation of parameters. First, need to solve the homogeneous equation (CHECK the coefficient of y is one). So,: Homogeneous Solution: y + 4y + 4y = 0 r 2 + 4r + 4 = 0 r = -2(multiplesolutions) yh = c1 e-2t + c2 te-2t . So, y1 = e-2t , y2 = te-2t Particular Solution: Let yp = u1 y1 + u2 y2 . Then: u1 = -y2 g(t) W (y1 ,y2 ) ; u2 = y1 g(t) W (y1 ,y2 ) . W (y1 , y2 ) = e-4t u1 = -(te-2t )(t-2 e-2t )/e-4t = -t-1 u1 = -ln(t) (No absolute sign as we consider only t > 0). u2 = -t-2 e-4t /e-4t = -t-2 u2 = t-1 . 5 So, the particular solution is yp = -ln(t)e-2t + y = yh + yp = c1 e-2t + c2 te-2t - ln(t)e-2t + (8) y + 4y = 3csc(2t) = g(t) Homogeneous Solution: te-2t t . So, the general solution is: te-2t t . y + 4y = 0 r 2 + 4 = 0 r = 2i y1 = cos(2t), y2 = sin(2t) yh = c1 cos(2t) + c2 sin(2t). Particular Solution: yp = u1 y1 + u2 y2 with: u1 = -y2 g(t) W (y1 ,y2 ) ; u2 = y1 g(t) W (y1 ,y2 ) . W (y1 , y2 ) = 2(cos(2t)2 - (-sin(2t)2 )) = 2(cos(2t)2 + sin(2t)2 ) = 2 u1 = -sin(2t)3csc(2t)/2 = -1/2 u1 = -t/2 u2 = cos(2t)csc(2t)/2 = cot(2t)/2 u1 = ln(sin(2t)) 4 (No absolute sign since 0 < t < /2). So, the particular solution is yp = -tcos(2t)/2+sin(2t) ln(sin(2t)) . So, the general solution is: 4 y = yh + yp = c1 cos(2t) + c2 sin(2t) - tcos(2t)/2 + sin(2t) ln(sin(2t)) . 4 (9) 4y + y = 2sec(t/2) First, need to make the coefficient of y one y + Homogeneous Solution: y 4 = sec(t/2) 2 = g(t) y + y = 0 r 2 + 1 = 0 r = i/2 y1 = cos(t/2), y2 = sin(t/2) yh = 4 4 c1 cos(t/2) + c2 sin(t/2). Particular Solution: yp = u1 y1 + u2 y2 u1 = -y2 g(t) W (y1 ,y2 ) ; u2 = y1 g(t) W (y1 ,y2 ) . W (y1 , y2 ) = 1/2. 6 u1 = -sin(t/2)sec(t/2) u1 = 2ln(cos(t/2)). u2 = cos(t/2)sec(t/2) u2 = t. So, the particular solution is yp = 2ln(cos(t/2))cos(t/2) + tsin(t/2). So, the general solution is: y = yh + yp = c1 cos(t/2) + c2 sin(t/2) + 2ln(cos(t/2))cos(t/2) + tsin(t/2). (13) t2 y - 2y = 3t2 - 1, t > 0, y1 = t2 , y2 = t-1 . Divide the whole equation with t2 to get: y - t2 y = 3 - 2 Particular Solution: yp = u1 y1 + u2 y2 u1 = -y2 g(t) W (y1 ,y2 ) ; u2 1 t2 = g(t) = y1 g(t) W (y1 ,y2 ) . W (y1 , y2 ) = -3. u1 = t-1 (-1 + t-2 /3) = t-1 - t-3 /3 u1 = ln(t) + t-2 /6. u2 = t2 (-1 + t-2 /3) = -t2 + 1/3 u2 = -t3 /3 + t/3. So, the particular solution is yp = t2 ln(t) + 1/6 - t2 /3 + 1/3 = t2 ln(t) + 1/2 - t2 /3. PAGE: 259 NOTE: In all power series problems, since we are solving a 2nd order ODE, it is assumed that a0 , a1 are arbitrary. They are like c1 , c2 used in defining the homogeneous solution. (2) y - xy - y = 0, x0 = 0 Let y = an (x - x0 )n = an xn . n=0 n=0 So, y = nan xn-1 n=1 y = n(n - 1)an xn-2 = (n + 2)(n + 1)an+2 xn . n=2 n=0 Plugging y, y , y into the original equation, 7 2a2 + (n + 2)(n + 1)an+2 xn - nan xn - a0 - an xn = 0 (2) n=1 n=1 n=1 a0 a2 = (3) 2 an (n + 2)(n + 1)an+2 - nan - an = 0 (n + 2)(n + 1)an+2 - (n + 1)an = 0 an+2 = (4) n+2 So, (n + 2)(n + 1)an+2 xn - x nan xn-1 - an xn = 0 (1) n=0 n=1 n=0 a1 a3 = 3 a2 a0 a4 = = 4 4.2 a3 a1 a5 = = 5 5.3 a4 a0 a6 = = 6 6.4.2 a1 a5 = a7 = 7 7.5.3 .. .. .. a0 a2n = n 2 n! 2n n!a1 = (2n + 1)! a2n+1 Plugging all the coefficients into the original equation, get: y = an xn = a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + a5 x5 + a6 x6 + a7 x7 + ... n=0 a1 a0 4 a1 5 a0 6 a1 7 a0 x + x + x + x + ... y = a0 + a1 x + x2 + x3 + 2 3 4.2 5.3 6.4.2 7.5.3 x2 x4 x6 x7 x3 x5 y = a0 (1 + + + + ...) +a1 (x + + + + ...) 2 8 48 3 15 105 y1 y2 y = a0 y1 + a1 y2 ; y1 = n=0 x2n 2n n! ; y 2 = n=0 x2n+1 2n n! (2n + 1)! 8 (3) y - xy - y = 0, x0 = 1 y - (x - 1 + 1)y - y = 0 y - (x - 1)y - y - y = 0 Let y = an (x - 1)n n=0 So, y = nan (x - 1)n-1 n=1 y = n(n - 1)an (x - 1)n-2 = (n + 2)(n + 1)an+2 (x - 1)n . n=2 n=0 Plugging these into the second equation, get: inf ty (n+2)(n+1)an+2 (x-1)n - nan (x-1)n - nan (x-1)n-1 - an (x-1)n = 0 n=1 n=1 n=0 n=0 Re-writing the indices in the last but first term, (n + 2)(n + 1)an+2 (x - 1)n - nan (x - 1)n - (n + 1)an+1 (x - 1)n - an (x - 1)n = 0 (5) n=0 n=1 n=0 n=0 2a2 + (n + 2)(n + 1)an+2 (x - 1)n - nan (x - 1)n n=1 n=1 -a1 - (n + 1)an+1 (x - 1)n - a0 - an (x - 1)n = 0 n=1 n=1 2a2 - a0 - a1 = 0 a2 = a0 + a1 2 (6) (7) (n + 2)(n + 1)an+2 - nan - (n + 1)an+1 - an = 0 an+2 = an + an+1 ;n 1 n+2 This gives the recurrence relation. From here, we need to get the two linearly independent power series solutions. Since, y = a0 y1 + a1 y2 , in order to get y1 , one can find the series solution with a1 = 0 factoring out a0 from the resulting coefficients and similarly to get y2 , can set a0 = 0 and find the an s and plug them in the power series to get y2 after factoring a1 . So, setting a1 = 0 to get y1 , 9 a2 = a1 + a2 a0 = 3 2.3 a0 + a0 a3 + a2 2 a4 = = 6 4 4 a4 + a3 a5 = 5 a5 + a4 a6 = 6 a3 = = = = = a0 2 a0 6 a0 6 a0 3 a0 2 .. .. (8) 1 1 1 1 1 y1 = 1 + (x - 1)2 + (x - 1)3 + (x - 1)4 + (x - 1)5 + (x - 1)6 + ... 2 6 6 3 2 So, setting a1 = 0 to get y1 , a1 2 a1 + a2 3a1 a1 a3 = = = 3 2.3 2 a3 + a2 a1 a4 = = 4 4 3a1 a4 + a3 = a5 = 5 10 a5 + a4 13a1 a6 = = 6 10 .. a2 = .. 1 1 1 3 13 y1 = (x - 1) + (x - 1)2 + (x - 1)3 + (x - 1)4 + (x - 1)5 + (x - 1)6 + ... 2 2 4 10 10 (5) (1 - x)y + y = 0, x0 = 0 Let y = an (x - x0 )n = an xn . n=0 n=0 So, y = nan xn-1 n=1 y = n(n - 1)an xn-2 = n(n + 1)an+1 xn-1 = (n + 2)(n + 1)an+2 xn . n=0 n=1 n=2 10 (9) Here, we have two definitiions for y as the given equation (1-x)y +y = 0 y -xy +y = 0 and for the first term use, y = (n + 2)(n + 1)an+2 xn and the for the second term use, n=0 y = n(n + 1)an+1 xn-1 as this term is multiplied by x resulting in a series with xn but n=1 starting from a different index. (n + 2)(n + 1)an+2 xn - n(n + 1)an+1 xn + an xn = 0 n=0 n=1 n=0 Collecting coefficients of powers of x: 2a2 + a0 + [(n + 2)(n + 1)an+2 - n(n + 1)an+1 + an ]xn = 0 n=1 So, 2a2 + a0 = 0 a2 = - a0 and (n + 2)(n + 1)an+2 - n(n + 1)an+1 + an = 0 which 2 gives the following recurrence relation n 1: an+2 = n(n + 1)an+1 - an (n + 2)(n + 1) In orfder to find the two linearly independent solutions y1 , y2 , follow the same procedure as before. y1 : Set a1 = 0 and solve for all the coefficients an . Then factor out the a0 i.e., ignore it and then assemble the whole series to get y1 -a0 2 2a2 - a1 -a0 a3 = = 6 6 6a3 - a2 -a0 a4 = = 12 24 . a2 = . . So, 1 1 1 y1 = 1 - x2 - x3 - x4 + ... 2 6 24 y2 : Set a0 = 0 and solve for all the coefficients an . Then factor out the a1 i.e., ignore it 11 and assemble then the whole series to get y2 -a0 =0 2 2a2 - a1 -a1 a3 = = 6 6 6a3 - a2 -a1 a4 = = 12 12 -a1 12a4 - a3 = a5 = 20 24 . a2 = . So, 1 1 1 y2 = x - x3 - x4 - x5 + ... 6 12 24 (7) y + xy + 2y = 0, x0 = 0 Let y = an (x - x0 )n = an xn . n=0 n=0 So, y = nan xn-1 n=1 y = n(n - 1)an xn-2 = (n + 2)(n + 1)an+2 xn . n=0 n=2 Plugging these expressions into the ODE: (n + 2)(n + 1)an+2 xn + x nan xn-1 + 2 an xn = 0 n=0 n=1 n=0 (n + 2)(n + 1)an+2 xn + nan xn + 2 an xn = 0 n=0 n=1 n=0 Gathering the n = 0 terms from the first and the last terms and collecting the rest of the terms into the summation: (2a2 + a0 )x0 + [(n + 2)(n + 1)an+2 + nan + 2an ]xn = 0 n=1 Setting the coefficients of powers of x equal to zero: 2a2 + a0 = 0 a2 = -a0 2 -an (n + 2) (n + 2)(n + 1) (n + 2)(n + 1)an+2 + nan + 2an = 0 an+2 = an+2 = -an ,n 1 n+1 Last expression gives the recurrence relation. It can be seen that the even coefficients depend only on past even coefficients and odd coefficients depend only on past odd coefficients. So, plugging in even values of n: 12 a2.3 -a0 2 -a2 2 2.a0 a2.2 = = (-1) 3 2.3! 4.2.a0 22 2!a0 -a4 = (-1)3 = (-1)3 = 5 2.5! 2.5! -a6 23 3!a0 a2.4 = = (-1)4 7 2.7! . a2 = . a2k = (-1)k 2k-1 (k - 1)!a0 2.(2k - 1)! So, So, plugging in odd values of n: 2k-1 (k - 1)! 2k 2 4 1 x + ... + (-1)k x + ... y1 = 1 - x2 + 2 2.3! 2.(2k - 1)! a2.1+1 = a2.2+1 = a2.3+1 a1 -a3 = (-1)2 4 4.2 -a5 3 a1 = (-1) = 6 6.4.2 -a7 a2.4+1 = 8 -a1 2 a1 = (-1)2 2 2 2! 3 a1 = (-1) 3 2 3! 4 a1 = (-1) 4 2 4! . a2k+1 So, . a1 = (-1)k k 2 k! 1 1 1 y2 = x - x3 + 2 x5 + ... + (-1)k k x2k+1 + ... 2 2 .2! 2 k! (9) (1 + x2 )y - 4xy + 6y = 0, x0 = 0 Let y = an (x - x0 )n = an xn . n=0 n=0 So, y = nan xn-1 n=1 13 y = n(n - 1)an xn-2 = (n + 2)(n + 1)an+2 xn . n=0 n=2 Plugging these expressions into the original ODEy + x2 y - 4xy + 6y = 0: (n + 2)(n + 1)an+2 xn + x2 n(n - 1)an xn-2 - 4x nan xn-1 + 6 an xn = 0 n=0 n=2 n=1 n=0 (n + 2)(n + 1)an+2 xn + n(n - 1)an xn - 4nan xn + 6an xn = 0 n=0 n=2 n=1 n=0 Gathering the n = 0 terms from the first and the last terms; n = 1 from first, third and last terms and collecting the rest of the terms into the summation: (2a2 + 6a0 )x0 + (6a3 - 4a1 + 6a1 )x + [(n + 2)(n + 1)an+2 + n(n - 1)an - 4nan + 6an ]xn = 0(10) n=2 Equating the coefficients of powers of x to zero: 2a2 + 6a0 = 0 a2 = -3a0 1 6a3 - 4a1 + 6a1 = 0 a3 = - a1 3 (n + 2)(n + 1)an+2 + n(n - 1)an - 4nan + 6an = 0 an+2 = -an (n2 - 5n + 6) -an (n - 2)(n - 3) = ;n 2 (n + 2)(n + 1) (n + 1)(n + 2) (11) (12) The last expression gives the recurrence relation. It can be seen that the even coefficients depend only on past even coefficients and odd coefficients depend only on past odd coefficients. So, plugging in even values of n: -a2 (2 - 2)(2 - 3) =0 4.3 -a4 (4 - 2)(4 - 3) a6 = =0 (4 + 1)(4 + 2) . a4 = . a2k = 0; k 2 So: For odd values of n: a5 = y1 = 1 - 3x2 -a3 (3 - 2)(3 - 3) =0 4.5 -a5 (5 - 2)(5 - 3) a6 = =0 (5 + 1)(5 + 2) . . a2k+1 = 0; k 2 14 (18) (19) (20) (21) (22) (13) (14) (15) (16) (17) So: 1 y 2 = x - x3 3 (10) (4 - x2 )y + 2y = 0, x0 = 0 Let y = an (x - x0 )n = an xn . n=0 n=0 So, y = nan xn-1 n=1 y = n(n - 1)an xn-2 = (n + 2)(n + 1)an+2 xn . n=0 n=2 Plugging these series in to the original ODE 4y - x2 y + 2y = 0: 4 (n + 2)(n + 1)an+2 xn - n(n - 1)an xn + 2 an xn = 0 n=0 n=2 n=0 Separating initial parts of series for n = 0, n = 1 and gathering the rest of the terms into a summation: (4.2a2 + 2a0 )x0 + (4.3.2a3 + 2a1 )x1 + [4(n + 2)(n + 1)an+2 - n(n - 1)an xn + 2an ]xn = 0 n=2 Setting the coefficients of all powers of x to zero: 2.a0 4.2 2a1 4.3.2a3 + 2a1 = 0 a3 = - 4.3.2 4.2a2 + 2a0 = 0 a2 = - 4(n + 2)(n + 1)an+2 - n(n - 1)an xn + 2an = 0 an+2 = an+2 (n2 - n - 2)an 4(n + 2)(n + 1) (n - 2)an ;n 2 = 4(n + 2) The last expression gives the recurrence relation. It can be seen that the even coefficients depend only on past even coefficients and odd coefficients depend only on past odd coefficients. So, plugging in even values of n: (2 - 2)a2 =0 4(2 + 2) (4 - 2)a4 =0 a6 = 4(4 + 2) . a4 = . a2k = 0; k 2 15 So, 1 y1 = 1 - x2 4 Plugging in odd values of n: a5 = 3a5 47 5a7 a9 = 49 a7 = a1 a3 =- 2 4.5 4 .5.3 2 a1 = (-1) 3 4 .7.5 a1 = (-1)3 4 4 .9.7 . . a2k+1 = (-1)k-1 So, y2 = x - PAGE 313: a1 4k (2k + 1)(2k - 1) 1 3 1 x + .. + (-1)k-1 k x2k+1 + .. 12 4 (2k + 1)(2k - 1) (15) Find Laplace Transform of teat L[teat ] = at -st dt 0 te e = -(s-a)t dt 0 te Integrating by parts: L[teat ] = at -st dt 0 te e = te-(s-a)t -(s-a) |0 + e-(s-a)t | (s-a)2 0 = -1 (s-a)2 (16) Find the Laplace Transform of tsin(at) L[teat ] = -st dt 0 tsin(at)e =t -st dt 0 sin(at)e Integrating by parts and using the following formulae: A eAt sin(Bt)dt = [ A2 +B 2 sin(Bt) - A eAt cos(Bt)dt = [ A2 +B 2 cos(Bt) + B cos(At)]eBt A2 +B 2 B sin(At)]eBt A2 +B 2 Simplify to get the answer. PAGE 322: 16 2 (1) L-1 [ s23 ] = 3 L-1 [ s2 +22 ] = 2 +4 3sin(2t) 2 1 s+4 )] 2 1 2 2 (3) L-1 [ s2 +3s-4 ] = L-1 [ (s+4)(s-1) ] = L-1 [ 5 ( s-1 - 2(s+1) (5) L-1 [ s22s+2 ] = L-1 [ (s+1)2 +22 ] = 2e-t cos(2t) +2s+5 = 2 [et - e-4t ] 5 2(s-1) (7) L-1 [ s22s+1 ] = L-1 [ 2(s-1)+3 ] = L-1 [ (s-1)2 +1 + -2s+2 (s-1)2 +1 -4s+12 (8) L-1 [ 8ss(s2 +4) ] = L-1 [ 8(s 2 2 +4)-4s-20 2 3 ] (s-1)2 +1 = et [2cos(t) + 3sin(t)] - 20 ] s(s2 +4) = 8- 3 - 2sin(2t) + 5cos(2t) 20L-1 [ 1 ] s(s2 +4) s(s2 +4) 2sin(2t) - 20 L-1 [ 1 4 s +4) ] = L-1 [ 8(s2 +4) - s(s 4s s(s2 +4) - s ] (s2 +4) = 8 - 2cos(2t) - 5[1 - cos(2t)] = = e-2t [5sin(t) - 2cos(t)] = 8 - 2sin(2t) - 5 (9) L-1 [ s21-2s ] = L-1 [ 5-2(s+2) ] = L-1 [ (s+2)2 +1 - +4s+5 (s+2)2 +1 2(s+2) ] (s+2)2 +1 2(s+1)-5 2(s+1) 2s-3 2s-3 5 (10) L-1 [ s2 +2s+10 ] = L-1 [ (s+1)2 +9 ] = L-1 [ (s+1)2 +32 ] = L-1 [ (s+1)2 +32 - (s+1)2 +32 ] = e-t [ 2 cos(3t)- 3 5 3 sin(3t)] Solve the following initial value problems using Laplace Transforms: (12) L[y + 3y + 2y] = 0 with y(0) = 1, y (0) = 0 i.e., 0 = L[y + 3y + 2y] = L[y](s2 + 3s + 2) - s - 3 s+3 L[y] = 2 s + 3s + 2 s+3 = (s + 1)(s + 2) 2 1 + = - (s + 2) (s + 1) y = -e-2t + 2e-t (13) L[y - 2y + 2y] = 0 with y(0) = 0, y (0) = 1 i.e., 17 = s(sL[y]) - 1) + 3L[y] - 3 + 2L[y] = s(sL[y]) - y(0)) - y (0) + 3(L[y] - y(0)) + 2L[y] = s(L[y ]) - y (0) + 3(L[y] - y(0)) + 2L[y] = L[y ] + 3L[y ] + 2L[y] 0 = L[y - 2y + 2y] = L[y ] - 2L[y ] + 2L[y] = L[y](s2 - 2s + 2) - 1 1 L[y] = 2 s - 2s + 2 1 = (s - 1)2 + 1 y = et sin(t) = s(sL[y])) - 1 - 2sL[y] + 2L[y] = s(sL[y]) - y(0)) - y (0) - 2(sL[y] - y(0)) + 2L[y] = s(L[y ]) - y (0) - 2(sL[y] - y(0)) + 2L[y] (14) L[y - 4y + 4y] = 0 with y(0) = 1, y (0) = 1 i.e., 0 = L[y - 4y + 4y] = L[y ] - 4L[y ] + 4L[y] = L[y](s2 - 4s + 4) - s + 3 s-3 L[y] = 2 s - 4s + 4 s-3 = 2 s - 4s + 4 (s - 2) - 1 = (s - 2)2 (s - 2) 1 = - 2 (s - 2) (s - 2)2 1 1 - = (s - 2) (s - 2)2 y = e2t - te2t (15) L[y - 2y + 4y] = 0 with y(0) = 2, y (0) = 0 i.e., 18 = s(sL[y])) - 1 - s - 4sL[y] + 4 + 4L[y] = s(sL[y]) - y(0)) - y (0) - 4(sL[y] - y(0)) + 4L[y] = s(L[y ]) - y (0) - 4(sL[y] - y(0)) + 4L[y] 0 = L[y - 2y + 4y] = L[y ] - 2L[y ] + 4L[y] = L[y](s2 - 2s + 4) - 2s + 4 2s - 4 L[y] = 2 s - 2s + 4 2s - 4 = (s - 1)2 + 3 2(s - 1) - 2 = (s - 1)2 + 3 2 2(s - 1) - = 2 + ( 3)2 2 + ( 3)2 (s - 1) (s - 1) 1 y = 2et [cos( 3t) - sin( 3t)] 3 = s(sL[y])) - 2s - 2sL[y] + 4 + 4L[y] = s(sL[y]) - y(0)) - y (0) - 2(sL[y] - y(0)) + 4L[y] = s(L[y ]) - y (0) - 2(sL[y] - y(0)) + 4L[y] PAGE 330: Find the inverse Laplace Transform of the following functions: (15) F (s) = 2(s-1)e-2s s2 -2s+2 = e-2s H(s) where H(s) = 2(s-1) s2 -2s+2 = 2(s-1) (s-1)2 +1 Now, L-1 [H(s)] = h(t) = 2et cos(t). So, the inverse Laplace transform of F (s) i.e.,. L-1 [e-2sH(s) ] = u2 (t)h(t - 2) = 2u2 (t)et-2 cos(t - 2) (16) F (s) = 2e-2s s2 -4 = e-2s H(s) where H(s) = 2 s2 -4 1 1 = 2 [ s-2 - 1 s+2 ] 1 Now, L-1 [H(s)] = h(t) = 2 [e2t - e-2t ]. So, the inverse Laplace transform of F (s) i.e.,. 2(t-2) -e-2(t-2) L-1 [e-2sH(s) ] = u2 (t)h(t - 2) = u2 (t) e = u2 (t)sinh(2(t - 2)) [Recalling that 2 t - sinh(t) = e -e t ] 2 (17) F (s) = (s-2)e-s s2 -4s+3 = e-s H(s) where H(s) = (s-2) s2 -4s+3 = (s-2) (s-2)2 -1 Now, L-1 [H(s)] = h(t) = e2t sinh(t). So, the inverse Laplace transform of F (s) i.e.,. L-1 [e-sH(s) ] = u1 (t)h(t - 1) = u1 (t)e2(t-1) cosh(t - 1) 19 (18) F (s) = e-s +e-2s -e-3s -e-4s s = [e-s + e-2s - e-3s - e-4s ] L[1]. This means that with H(s) H(s) = L[h(t)], have h(t) = 1. Now L-1 [F (s)] = u1 (t)h(t - 1) + u2 (t)h(t - 2) - u3 (t)h(t - 3) - u4 (t)h(t - 4) = u1 (t) + u2 (t) - u3 (t) - u4 (t) PAGE 337: (1) y + y = f (t) with y(0) = 0, y (0) = 1 and f (t) = u0 (t) - u/2 (t). Taking Laplace Transform of the equation with Y (s) = L[y]: L[f (t)] = 1 - e-/2s s = L[y ] + L[y] = s(sL[y] - y(0)) - y (0) + L[y] 1 - s - e-/2s s(s2 + 1) = sL[y ] - y (0) + L[y] = (s2 + 1)L[y] - 1 Y (s) = = y(t) = 1 - cos(t) - sin(t) - u/2 (t)[1 - sin(t)] (2) y + 2y + 2y = h(t) with y(0) = 0, y (0) = 1 and h(t) = u (t) - u2 (t) Taking Laplace Transform of the equation with Y (s) = L[y]: 1-s e-/2s - s(s2 + 1) s(s2 + 1) s 1 1 s 1 - 2 - 2 - e-/2s [ - 2 ] = s s +1 s +1 s (s + 1) y(t) = 1 - cos(t) - sin(t) - u/2 (t)[1 - cos(t - /2)] 20 L[h(t)] = e-s - e-2s s = L[y ] + 2L[y ] + 2L[y] = s(sY (s) - y(0)) - y (0) + 2sY (s) - 2y(0) + 2Y (s) = sL[y ] - y (0) + 2sY (s) - 2y(0) + 2Y (s) = s2 Y (s) - 1 + 2sY (s) + 2Y (s) e-s - e-2s + s Y (s) = s(s2 + 2s + 2) e-s - e-2s + s = s((s + 1)2 + 1) e-2s 1 e-s - + = 2 + 1) 2 + 1) s((s + 1) s((s + 1) (s + 1)2 + 1 y(t) = u (t)g(t - ) - u2 (t)g(t - 2) + e-t sin(t) where G(s) = L[g(t)] = 1 s((s+1)2 +1) (23) = 1[1 - 2 s 1 (s+1)2 +1 - s+1 (s+1)2 +1 ]. So: g(t) = 1 [1 - e-t sin(t) - e-t cos(t)] 2 So, y(t) = u2(t) [1 - e-(t-) sin(t - ) - e-(t-) cos(t - )] - u2 (t) [1 - e-(t-2) sin(t - 2) - 2 e-(t-2) cos(t - 2)] + e-t sin(t). Simplifying this: y(t) = u2(t) [1 + e-(t-) sin(t) + e-(t-) cos(t)] - e-t sin(t) u2 (t) -(t-2) sin(t) - e-(t-2) cos(t)] + 2 [1 - e (3) y + 4y = sin(t) + u2 (t)sin(t - 2) with y(0) = 0, y (0) = 0. Taking Laplace Transform of the equation with Y (s) = L[y]: 21 L[f (t)] = 1 + e-2s s2 + 1 = L[y ] + 4L[y] = sL[y ] - y (0) + 4Y (s) = s(sY (s) - y(0)) - y (0) + 4Y (s) = (s2 + 4)Y (s) 1 + e-2s Y (s) = (s2 + 4)(s2 + 1) 1 1 (1 + e-2s ) [ 2 - 2 ] = 3 s +1 s +4 1 1 1 e-2s 1 1 = [ 2 - 2 ]+ [ 2 - 2 ] 3 s +1 s +4 3 s +1 s +4 1 u2 (t) 1 1 [sin(t) - sin(2t)] + [sin(t - 2) - sin(2t - 4)] y(t) = 3 2 3 2 (7) y + y = u3 (t) with y(0) = 1, y (0) = 0. Taking Laplace Transform of the equation with Y (s) = L[y]: L[u3 (t)] = e-3s s = L[y ] + L[y] = sL[y ] - y (0) + Y (s) = s(sY (s) - y(0)) - y (0) + Y (s) = (s2 + 1)Y (s) - s e-3s + s2 Y (s) = s(s2 + 1) e-3s s = + 2 2 + 1) s(s s +1 = e-3s [1/s - s/(s2 + 1)] + y(t) = u3 (t)[1 - cos(t)] + cos(t) s2 s +1 (8) y + y + 5y/4 = t - u/2 (t)(t - /2) with y(0) = 0, y (0) = 0. Taking Laplace Transform of the equation with Y (s) = L[y]: 22 L[t - u/2 (t)(t - /2)] = 1/s2 - e-/2s /s2 = L[y ] + L[y ] + 5/4L[y] = (s2 + s + 5/4)Y (s) = sL[y ] - y (0) + L[y ] + 5/4Y (s) e-/2s 1 - 2 (s2 + s + 5/4)s2 (s + s + 5/4)s2 (24) Y (s) = = [A/s + B/s2 + (Cs + D)/(s2 + s + 5/4)][1 - e/2s ] Pointers for Fourier Series questions: For a function f (x) periodic with a period 2L i.e., f (x + 2L) = f (x) a0 f (x) = + [an cos(nx/L) + bn sin(nx/L)] n=1 2 Euler - Fourier Formulae: For n 1 an = bn = For f (x) even i.e., f (-x) = f (x): a0 = For n 1 an = bn 2 L = 0 0 a0 = 1 L L L f (x)dx -L 1 L 1 L f (x)cos(nx/L)dx -L L f (x)sin(nx/L)dx -L 2 L L L f (x)dx 0 f (x)cos(nx/L)dx For f (x) odd i.e., f (-x) = -f (x): For n 1 an = 0 2 bn = L a0 = 0 L f (x)sin(nx/L)dx 0 23 A function f (x) can also be neither odd nor even in which case all the coefficients need to be computed. PLEASE DO THE REVIEW EXAMS POSTED ONLINE ON DR. WILTON'S WEBSITE. ALSO USE THE EXAMPLE SOLUTIONS AS A GUIDE. USE THEM TO SEE HOW TO SOLVE A PARTICULAR PROBLEM. ALL THE BEST FOR YOUR FINALS!! 24
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Gaspar, Adrian Exam 1 Due: Oct 2 2007, 11:00 pm Inst: MC Caputo This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part
University of Texas - M - 408L
Gaspar, Adrian Homework 1 Due: Sep 4 2007, 3:00 am Inst: MC Caputo This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (pa
University of Texas - CH - 302
amg2845 hw06 McCord (53580) This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Which of the five solutions below is most basic? 1. 0.10 M C
University of Texas - CH - 302
amg2845 hw05 McCord (53580) This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points The term &quot;Ka for the ammonium ion&quot; describes the equilibrium
University of Texas - CH - 302
Version 147 hw02 McCord (53580) This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Calculate the molality of sucrose in a solution composed
University of Texas - CH - 302
Chapter 6 Chemical Equilibrium Introduction o Chemical equilibrium- the state in which the concentrations of all reactants and products remain constant with time. o Any chemical reaction carried out in a closed vessel will reach equilibrium. 6.1 The
University of Texas - CH - 302
amg2845 Exam 1 McCord (53580) This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 1 atm = 760 torr 1 atm = 101325 Pa 1 bird/hand = 2 bird/bush 001 10.0 poin
University of Texas - CH - 302
Version 169 Exam 2 McCord (53580) This print-out should have 27 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. McCord for 1pm 5. 31 C 6. 135 C Explanation: T1 = 25 C +273 = 298 K H
University of Texas - CH - 302
Version 147 hw01 McCord (53580) This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. Here are some questions to get you thinking quantitatively again. Review
University of Texas - CH - 302
amg2845 hw04 McCord (53580) This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points What is the hydroxide concentration [OH- ] in an aqueous sol
University of Texas - CH - 302
amg2845 hw03 McCord (53580) This print-out should have 31 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. NOTE: If you see a number with an e in it followed by a signed number, you n
University of Texas - CH - 302
Dr. McCord CH302OVERVIEW for EXAM 1page 1Textbook Chapters for Exam 1Chapter 10 sections 9-11. Chapter 16 sections 10 &amp; 11. All of Chapter 17.Vapor PressureBe sure you understand what vapor pressure is. Why do we even say vapor? Is there an
University of Texas - ITL - 506
[k]ca checo chi ci cioculike catITL506 CAPRA[t] ciacelike cherry ciuLab Manual pp 208-209CD 1; track 25[g]ga ghego ghigulike go[d ] giage gi gio giulike JerryLM pp 229-230CD 3; track 24[]sce sci scia scio s
University of Texas - ITL - 506
ESERCIZIO - PreposizioniCapra1. Tutti I weekend viaggio _ Austin _ Dallas, ma io sono _ San Antonio. 3. - _ chi e questo libro? - _ Laura. 4. - _ chi sono questi cioccolatini? - Sono per la nonna. 5. Ieri abbiamo studiato _ 6:00 _ 9:00. 6. - _ ch
University of Texas - BIO - 311C
Worksheet: Transcription &amp; Translation of the Genetic Code Part I: Transcribing the Genetic Code (DNA mRNA) DNA sequence that will be transcribed to make messenger RNA: #1 #2Chapter 17 (Buskirk)G T T A C C T C G A TA G T C A T A A G A T T C G A
University of Texas - BIO - 311C
Chapter 1- Exploring Life Biological Hierarchy- from the smallest building blocks to the whole picture. o Molecules-&gt; Organelles-&gt; Cells-&gt; Tissues-&gt; Organs/Organ Systems-&gt; Organisms-&gt; Populations-&gt; Communities-&gt; Ecosystems-&gt; The Biosphere Emergent Pr
University of Texas - ITL - 506
ESERCIZI Verbi andare via - uscire - partire - lasciare - dire - parlare 1. Sono stanco, devo _ . 2. A che ora _ per Roma? 3. Con chi _ stasera? 4. Dopo che la lezione finisce, tutti _ . 5. Stamattina io _ I compiti a casa. 6. Stefano, con chi _? 7.