Answered: 9x 4. y" -9y=3x 5. 4y" — y=xe*¹/2, y(0)… | bartleby

Advanced Engineering Mathematics

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Solve the following ODEs of higher order variation of parameters. Help me with these 3 problems

9x
4. y"-9y=-
= 3x
5. 4y" - y = xe*/2², y(0) = 1,y'(0) = 0
2y"+y'-y = x + 1, y(0) = 1,y'(0) = 0
6.

Expert Solution

Step 1: Solving using variation of parameter

$y'' - 9 y = \frac{9 x}{e^{3 x}}$

Finding the C.F

$D^{2} - 9 = 0 \Rightarrow D = 3, - 3$

Therefore

$C . F = c_{1} e^{3 x} + c_{2} e^{- 3 x}$

P.I $= u_{1} (x) v_{1} (x) + u_{2} (x) v_{2} (x)$

$= v_{1} e^{3 x} + v_{2} e^{- 3 x}$

$W = |\begin{matrix} e^{3 x} & e^{- 3 x} \\ 3 e^{3 x} & - 3 e^{- 3 x} \end{matrix}| = - 6$

$v_{1} = - \int \frac{e^{- 3 x} 9 x e^{- 3 x}}{- 6} d x = \frac{3}{2} \int x e^{- 6 x} d x = \frac{3}{2} [- x \frac{e^{- 6 x}}{6} - \int \frac{d x}{d x} [\int e^{- 3 x} d x] d x] = [- \frac{x e^{- 6 x}}{4} - \frac{3}{2} \int \frac{e^{- 6 x}}{- 6} d x] = [- \frac{x e^{- 6 x}}{4} - \frac{1}{24} e^{- 6 x}]$

$v_{2} = \int \frac{e^{3 x} 9 x e^{- 3 x}}{- 6} = - \frac{3}{2} \int x d x = - \frac{3 x^{2}}{4}$

P.I $= - (\frac{x e^{- 6 x}}{4} + \frac{1}{24} e^{- 6 x}) e^{3 x} - \frac{3 x^{2}}{4} e^{- 3 x}$

$= - (\frac{x}{4} + \frac{1}{24}) e^{- 3 x} - \frac{3 x^{2}}{4} e^{- 3 x}$

Therefore

$y (x) = c_{1} e^{3 x} + c_{2} e^{- 3 x} - (\frac{x}{4} + \frac{1}{24}) e^{- 3 x} - \frac{3 x^{2}}{4} e^{- 3 x}$

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