[ (x+1)^(2 ) D^2 +(x+1)D+1 ]y = 4cos (ln( x+1))


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Question Number 130827 by bemath last updated on 29/Jan/21

$[{(x + 1)}^{2} D^{2} + (x + 1) D + 1] y = 4 \cos (\ln (x + 1))$
	Answered by EDWIN88 last updated on 29/Jan/21
	$let ln (x+1)=t ⇒x+1 = e^t { (((dy/dx) = (dy/dt)×(dt/dx) = (1/(x+1)).(dy/dt))),(((d^2 y/dx) = (d/dx) [ (1/(x+1)) (dy/dt) ]= (1/((x+1)^2 )) [(d^2 y/dt^2 )−(dy/dt) ])) :} ((d^2 y/dt^2 )−(dy/dt))+(dy/dt) +y = 4cos t (d^2 y/dt^2 ) +y = 4cos t for homogenous solution y_h = C_1 cos t + C_2 sin t particular solution y_p = At cos t + Bt sin t we get { ((A=0)),((B=2)) :} ⇒y_p = 2t sin t General solution y = C_1 cos t + C_2 sin t + 2t sin t y = C_1 cos (ln (x+1))+C_2 sin (ln (x+1))+2ln (x+1)sin (ln (x+1))$
	$l e t \ln (x + 1) = t \Rightarrow x + 1 = e^{t}$ ${\begin{cases} \frac{d y}{d x} = \frac{d y}{d t} \times \frac{d t}{d x} = \frac{1}{x + 1} . \frac{d y}{d t} \\ \frac{d^{2} y}{d x} = \frac{d}{d x} [\frac{1}{x + 1} \frac{d y}{d t}] = \frac{1}{{(x + 1)}^{2}} [\frac{d^{2} y}{{d t}^{2}} - \frac{d y}{d t}] \end{cases}$ $(\frac{d^{2} y}{{d t}^{2}} - \frac{d y}{d t}) + \frac{d y}{d t} + y = 4 \cos t$ $\frac{d^{2} y}{{d t}^{2}} + y = 4 \cos t$ $f o r h o m o g e n o u s s o l u t i o n$ $y_{h} = C_{1} \cos t + C_{2} \sin t$ $p a r t i c u l a r s o l u t i o n$ $y_{p} = A t \cos t + B t \sin t$ $w e g e t {\begin{cases} A = 0 \\ B = 2 \end{cases} \Rightarrow y_{p} = 2 t \sin t$ $G e n e r a l s o l u t i o n$ $y = C_{1} \cos t + C_{2} \sin t + 2 t \sin t$ $y = C_{1} \cos (\ln (x + 1)) + C_{2} \sin (\ln (x + 1)) + 2 \ln (x + 1) \sin (\ln (x + 1))$
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