solve-the-diff-eq-y-y-4y-4y-e-x-

Question Number 91010 by jagoll last updated on 27/Apr/20

s o l v e t h e d i f f e q

y^{‴} - y^{″} + 4 y^{'} - 4 y = e^{x}

Commented by john santu last updated on 27/Apr/20

t h e c h a r a c t e r i s t i c e q u a t i o n

w^{3} - w^{2} + 4 w - 4 = 0

w i t h r o o t s a r e w = 1, \pm 2 i

c o m p l e m e n t a r y s o l u t i o n

y_{c} = K_{1} e^{x} + K_{2} \cos 2 x + K_{3} \sin 2 x

p a r t i c u l a r s o l u t i o n

y_{p} = {Q x e}^{x} + {R e}^{x}

y^{'} = {Q x e}^{x} + (Q + R) e^{x}

y^{″} = {Q x e}^{x} + (2 Q + R) e^{x}

y^{‴} = {Q x e}^{x} + (3 Q + R) e^{x}

c o m p a r i n g c o e f f

y^{‴} - y^{″} + 4 y^{'} - 4 y = 5 {Q e}^{x} = e^{x}

Q = \frac{1}{5}

g e n e r a l s o l u t i o n

y = \frac{1}{5} {x e}^{x} + K_{1} e^{x} + K_{2} \cos 2 x + K_{3} \sin 2 x

Commented by jagoll last updated on 27/Apr/20

t h a n k y o u b o t h

Answered by MWSuSon last updated on 27/Apr/20

A u x i l l a r y e q u a t i o n

m^{3} - 4 m^{2} + 4 m - 4 = 0

(m - 1) (m^{2} + 4) = 0

m = 1, \pm i 2

y_{c} = C_{1} e^{x} + C_{2} \cos (2 x) + C_{3} \sin (2 x)

y_{p} = \frac{1}{D^{3} - D^{2} + 4 D - 4} e^{x}

= \frac{1}{1 - 1 + 4 - 4} e^{x} w h i c h i s u n d e f i n d e d

s o y_{p} = x \frac{1}{3 D^{2} - 2 D + 4} e^{x}

y_{p} = x \frac{1}{3 - 2 + 4} e^{x} = \frac{{x e}^{x}}{5}

y = C_{1} e^{x} + C_{2} \cos (2 x) + C_{3} \sin (2 x) + \frac{{x e}^{x}}{5}

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