Assignment # 8, Math 370, Fall 018 SOLUTIONS: Problem 1: Solve the equations (a) y y = 3x + x 4, (i) y(0) = 1, y (0) = 1, y (0) = 1. Characteristic equation: α 3 α = 0 so α 1, = 0 and α 3 =. y c = C 1 + C x + C 3 e x. Particular solution of non-homogeneous equation: normally we would look for y p in the form y p = Ax + Bx + C. Since 0 is a double root of the characteristic equation (0 corresponds to constant and polynomial solutions of homogeneous equation) we have to multiply y p by x. We take Then y p = Ax 4 + Bx 3 + Cx. y p = 4Ax 3 + 3Bx + Cx, y p = 1Ax + 6Bx + C, y p = 4Ax + 6B. We substitute and obtain which gives 4Ax + 6B 4Ax 1Bx 4C = 3x + x 4, 4A = 3 4A 1B = 1 6B 4C = 4, which gives A = 1, B = 8 1, C = 1. The general solution is 3 y = C 1 + C x + C 3 e x 1 8 x4 1 3 x3 + 1 x. IVP: y(0) = 1 gives 1 = C 1 + C 3. y = C + C 3 e x 1 x3 x + x, so y (0) = 1 gives 1 = C + C 3. y = 4C 3 e x 3 x x + 1, so y (0) = 1 gives 1 = 4C 3 + 1. The system of equations gives C 3 = 0 and C 1 = C = 1. The solution of IVP is y = 1 + x 1 8 x4 1 3 x3 + 1 x. (b) y 3y y = 4xe x, (i) y(0) = 1, y (0) = 1, y (0) = 1. Characteristic equation: α 3 3α = 0 so α 1, = 1 and α 3 =. y c = C 1 e x + C xe x + C 3 e x.
Particular solution of non-homogeneous equation: normally we would look for y p in the form y p = (Ax + B)e x. Since 1 is not a root of the characteristic equation (1 corresponds to solutions e x of homogeneous equation) we leave y p as it is. We take Then y p = (Ax + B)e x. y p = Aex +(Ax+B)e x = (Ax+A+B)e x, y p = (A)ex +(Ax+A+B)e x = (Ax+A+B)e x, We substitute and obtain which gives y p = (Ax + 3A + B)e x. e x (Ax + 3A + B 3Ax 3A 3B Ax B) = e x ( 4x), 4A = 4 4B = 0, which gives A = 1, B = 0. The general solution is y = C 1 e x + C xe x + C 3 e x + xe x. IVP: y(0) = 1 gives 1 = C 1 + C 3. y = C 1 e x C xe x +C e x +C 3 e x +xe x +e x, so y (0) = 1 gives 1 = C 1 +C +C 3 +1. y = C 1 e x + C xe x C e x C e x + 4C 3 e x + xe x + e x, so y (0) = 1 gives 1 = C 1 C + 4C 3 +. We have to solve C 1 + C 3 = 1 C 1 + C + C 3 = 0 C 1 C + 4C 3 = 1 The system of equations gives C 3 = 0 and C 1 = C = 1. The solution of IVP is y = e x + xe x + xe x. (c) y + 6y + 13y = e 3x cos x, (i) y(0) = 1, y (0) = 1. Characteristic equation: α + 6α + 13 = 0, = 36 5 = 16 so α 1, = 3 ± i. y c = C 1 e 3x cos(x) + C e 3x sin(x). Particular solution of non-homogeneous equation: normally we would look for y p in the form y p = (A cos x+b sin x)e 3x. Since 3 ±i is not a root of the characteristic equation ( 3 ±i corresponds to solutions e 3x cosx of homogeneous equation) we leave y p as it is. We take y p = (A cosx + B sinx)e 3x.
Then y p = e 3x ( 3A cos x 3B sinx A sinx + B cos x) = e 3x (( 3A + B)cosx + ( 3B A)sinx). y p = e 3x ((9A 3B)cos x + (9B + 3A)sinx + (3A B)sinx + ( 3B A)cosx) = e 3x ((8A 6B)cos x + (8B + 6A)sinx). We substitute and obtain e 3x ((8A 6B 18A + 6B + 13A)cos x + (8B + 6A 18B 6A + 13B)sin x) = e 3x cos x, which gives 3A = 1 3B = 0, which gives A = 1/3, B = 0. The general solution is y = C 1 e 3x cos(x)+c e 3x sin(x)+ 1 3 e 3x cosx = e 3x ( C 1 cos(x) + C sin(x) + 1 3 cos x ). IVP: y(0) = 1 gives 1 = C 1 + 1. 3 y = e 3x (( 3C 1 + C )cos(x) + ( 3C C 1 )sin(x) cosx + sinx), so y (0) = 1 gives 1 = 3C 1 + C 1. These give C 1 = /3 and C =. The solution of IVP is ( y = e 3x 3 cos(x) + sin(x) + 1 ) 3 cosx. Problem : Solve the equation y + 9y = 9e 3x + 18sin(3x) 9cos(3x), y(0) = 1, y (0) = 1, y (0) = 1. Characteristic equation: α 3 + 9α = 0 so α 1, = ±3i and α 3 = 0. y c = C 1 + C cos(3x) + C 3 sin(3x). Particular solution of non-homogeneous equation: we will do this in two steps, first for RHS 9e 3x and then for RHS 18sin(3x) 9cos(3x). (i) normally we would look for y p in the form y p = Ae 3x. Since 3 is not a root of the characteristic equation (3 corresponds to solutions e 3x of homogeneous equation) we leave y p as it is. We take Then y p = Ae 3x. y p = 3Ae 3x, y p = 9Ae 3x, y p = 7Ae 3x.
We substitute and obtain (7A + 7A)e 3x = 9e 3x, which gives A = 1/6. This part of particular solution is y p = 1 6 e3x. (ii) normally we would look for y p in the form y p = A cos(3x) + B sin(3x). Since ±3i is a single root of the characteristic equation we need to multiply this y p by x. We take Then y p = Ax cos(3x) + Bx sin(3x). y p = A cos(3x)+b sin(3x) 3Ax sin(3x)+3bx cos(3x) = (A+3Bx)cos(3x)+(B 3Ax)sin(3x), y p = 3B cos(3x) 3A sin(3x) 3(A + 3Bx)sin(3x) + 3(B 3Ax)cos(3x) = (6B 9Ax)cos(3x) + ( 6A 9Bx)sin(3x), y p = 9A cos(3x) 9B sin(3x) 3(6B 9Ax)sin(3x) + 3( 6A 9Bx)cos(3x) = ( 7A 7Bx)cos(3x) + ( 7B + 7Ax)sin(3x). We substitute and obtain ( 7A 7Bx)cos(3x) + ( 7B + 7Ax)sin(3x) + 9(A + 3Bx)cos(3x) + 9(B 3Ax)sin(3x) The terms with x on the LHS cancel and we have 18A = 9 18B = 18, which gives A = 1 The general solution is and B = 1. This part of particular solution is y p = 1 x cos(3x) x sin(3x). y = C 1 + C cos(3x) + C 3 sin(3x) 1 6 e3x + 1 x cos(3x) x sin(3x). IVP: y(0) = 1 gives 1 = C 1 + C 1 6. = 18sin(3x) 9cos(3x). y = 3C sin(3x) + 3C 3 cos(3x) 1 e3x + 1 cos(3x) 3 x sin(3x) sin(3x) 3x cos(3x), so y (0) = 1 gives 1 = 3C 3 1 + 1. y = 9C cos(3x) 9C 3 sin(3x) 3 e3x 3sin(3x) 9 x cos(3x) 6cos(3x) + 9x sin(3x), so y (0) = 1 gives 1 = 9C 3 6. We have to solve C 1 + C = 7 6 3C 3 = 1 9C = 17
The system of equations gives C 3 = 1, C 3 1 = 19 and C 9 = 17. The solution of IVP is 18 y = 19 9 17 18 cos(3x) + 1 3 sin(3x) 1 6 e3x + 1 x cos(3x) x sin(3x)x. Problem 3: Consider a general linear non-homogeneous DE with constant coefficients a n y (n) + a n 1 y (n 1) + + a 1 y + a 0 y = g(x). Below, You are given the roots of characteristic equation and the RHS g(x). Write the form of the particular solution y p You would use to find it by the Method of Undefined Coefficients. DO NOT SOLVE. (a) roots:,, 1 ; g(x) = x e x + xe x ; For g 1 (x) = x e x, standard is y p = (Ax + Bx + C)e x. Since α = corresponding to such y p is a double root of characteristic equation we have to multiply it by x. We use y p = x (Ax + Bx + C)e x. For g (x) = xe x, standard is y p = (Ax + B)e x. Since α = 1 corresponding to such y p is a single root of characteristic equation we have to multiply it by x. We use y p = x(ax + B)e x. (b) roots: 4, 0, + i, i; g(x) = x + sin(x) + x cos x; For g 1 (x) = x, standard is y p = (Ax + Bx + C). Since α = 0 corresponding to such y p is a single root of characteristic equation we have to multiply it by x. We use y p = x(ax + Bx + C). For g (x) = sin(x), standard is y p = A sin(x)+b cos(x). Since α = ±i corresponding to such y p is not a root of characteristic equation we keep it as it is. We use y p = A sin(x) + B cos(x). For g 3 (x) = x cos x, standard is y p = (Ax +Bx +C)sinx +(Ex +Fx + G)cos x. Since α = ±i corresponding to such y p is not a root of characteristic equation we keep it as it is. We use y p = (Ax + Bx + C)sinx + (Ex + Fx + G)cos x. (c) roots:, + 3i, 3i; g(x) = sin(3x)e x + cos(x)e x ; For g 1 (x) = sin(3x)e x, standard is y p = (A cos(3x) + B sin(3x))e x. Since α = ± 3i corresponding to such y p is a single root of characteristic equation we have to
multiply it by x. We use y p = x(a cos(3x) + B sin(3x))e x. For g (x) = cos(x)e x, standard is y p = (A cos x+b sinx)e x. Since α = ±i corresponding to such y p is not a root of characteristic equation we keep it as it is. We use y p = (A cosx + B sinx)e x. Problem 4: Solve equation: (a) y + 3y + y = e x + e x, y(0) = 0, y (0) = 0. Characteristic equation: α + 3α + = 0 so α 1 = 1 and α =. y c = C 1 e x + C e x. We will use the Method of Variation of Parameters: Since the equation is in standard form we do not have to standardize it. Let y 1 = e x and y = e x. Then, y 1 = e x and y = e x. We get the system of equations C 1y 1 + C y = 0 C 1y 1 + C y Adding both equation we obtain Then, or = e x +e x C 1e x + C e x = 0 C 1e x C e x C e x = e x + e x or C = ex + e x. = e x +e x C 1 = C e x = 1 + e x. We have C = ex + e xdx = ex = t, e x dx = dt} = 1 + t dt = ln( + t) = ln( + ex ) + D. 1 C 1 = + e xdx = ex = t, e x 1 dx = dt} = t( + t) dt = 1 ( 1 t 1 ) dt + t = 1 ( ) t ln + D 1 = 1 ( ) e x + t ln + D + e x 1 = 1 x ln( + ex ) + D 1. We obtain the general solution y = ( 1 x 1 ln( + ex ) + D 1 )e x + ( ln( + e x ) + D )e x. IVP: y(0) = 0 gives 0 = 1 ln3 + D 1 ln3 + D.
We have e x y = ( 1 1 ( 1 + e x)e x x 1 ln( + ex ) + D 1 )e x + ( ex ( ln( + e x ) + D + e x)e x )e x, so y (0) = 0 gives 0 = 1 1 1 3 1 1 + ln3 D 3. We have D 1 + D = 3 ln3 D 1 D = 5 ln3, which gives D 1 = 1 ln3 and D = ln3. The solution of IVP is y = ( 1 x 1 ln( + ex ) + 1 ln3)e x + ( ln( + e x ) + ln3)e x. (b) y + 4y = 4 sin(x), y(π/4) = 0, y (π/4) = 0. Characteristic equation: α + 4 = 0 so α 1, = ±i. y c = C 1 cos(x) + C sin(x). We will use the Method of Variation of Parameters: Since the equation is in standard form we do not have to standardize it. Let y 1 = cos(x) and y = sin(x). Then, y 1 = sin(x) and y = cos(x). We get the system of equations C 1y 1 + C y = 0 C 1y 1 + C y = 4 sin(x) or C 1 cos(x) + C sin(x) = 0 C 1 sin(x) + C cos(x) = 4 sin(x) We divide second equation by and multiply by sin(x). The first we multiply by cos(x). We add them and obtain Then from the first equation C 1 = sin(x) sin(x) = 4sin x cosx sin(x) = 4cos x. C = C 1 cos(x) sin(x) = 4cos x cos(x) sin x cosx = cos(x) sinx. We integrate C 1 = 4sinx + D 1.
cos(x) cos C = sin x dx = x sin x sin x 1 t = t = cos x, dt = sinxdx} = 1 t dt = ( = 1 ( 1 1 t + 1 1 + t = 4cos x + ln(1 cos x) ln(1 + cos x) + D. The general solution is cos x 1 dx = sin x sinxdx ( t 1 t 1 1 t )) dt = 4t + ln(1 t) ln(1 + t) + D ) dt = y = ( 4sinx + D 1 )cos(x) + (4cos x + ln(1 cos x) ln(1 + cos x) + D )sin(x). IVP: y(π/4) = 0 gives 0 = + ln(1 /) ln(1 + /) + D so D = ln(1 /) + ln(1 + /) = ( + ) + ln. We have y = ( 4cos x)cos(x) ( 4sin x + D 1 )sin(x) y (π/4) = 0 gives so sinx + ( 4sin x + (1 cos x) + sin x (1 + cosx) )sin(x) + (4cos x + ln(1 cos x) ln(1 + cos x) + D )cos(x). 0 = ( + D 1 ) + ( + D 1 =. / / (1 /) + (1 + /) ), The solution of the IVP is ( y = 4sin x + ( ) cos(x)+ 4cos x + ln(1 cos x) ln(1 + cosx) ( + )) + ln sin(x).