x-1-2-D-2-x-1-D-1-y-4cos-ln-x-1-

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

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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