Question Number 130827 by bemath last updated on 29/Jan/21
![[ (x+1)^(2 ) D^2 +(x+1)D+1 ]y = 4cos (ln( x+1))](https://www.tinkutara.com/question/Q130827.png)
$$\left[\:\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}\:} \mathrm{D}^{\mathrm{2}} +\left(\mathrm{x}+\mathrm{1}\right)\mathrm{D}+\mathrm{1}\:\right]\mathrm{y}\:=\:\mathrm{4cos}\:\left(\mathrm{ln}\left(\:\mathrm{x}+\mathrm{1}\right)\right) \\ $$
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))](https://www.tinkutara.com/question/Q130836.png)
$${let}\:\mathrm{ln}\:\left({x}+\mathrm{1}\right)={t}\:\Rightarrow{x}+\mathrm{1}\:=\:{e}^{{t}} \\ $$$$\begin{cases}{\frac{{dy}}{{dx}}\:=\:\frac{{dy}}{{dt}}×\frac{{dt}}{{dx}}\:=\:\frac{\mathrm{1}}{{x}+\mathrm{1}}.\frac{{dy}}{{dt}}}\\{\frac{{d}^{\mathrm{2}} {y}}{{dx}}\:=\:\frac{{d}}{{dx}}\:\left[\:\frac{\mathrm{1}}{{x}+\mathrm{1}}\:\frac{{dy}}{{dt}}\:\right]=\:\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:\left[\frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} }−\frac{{dy}}{{dt}}\:\right]}\end{cases} \\ $$$$ \\ $$$$\left(\frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} }−\frac{{dy}}{{dt}}\right)+\frac{{dy}}{{dt}}\:+{y}\:=\:\mathrm{4cos}\:{t} \\ $$$$\:\frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} }\:+{y}\:=\:\mathrm{4cos}\:{t}\: \\ $$$${for}\:{homogenous}\:{solution}\: \\ $$$${y}_{{h}} \:=\:{C}_{\mathrm{1}} \mathrm{cos}\:{t}\:+\:{C}_{\mathrm{2}} \mathrm{sin}\:{t} \\ $$$${particular}\:{solution}\: \\ $$$${y}_{{p}} \:=\:{At}\:\mathrm{cos}\:{t}\:+\:{Bt}\:\mathrm{sin}\:{t}\: \\ $$$${we}\:{get}\:\begin{cases}{{A}=\mathrm{0}}\\{{B}=\mathrm{2}}\end{cases}\:\:\Rightarrow{y}_{{p}} =\:\mathrm{2}{t}\:\mathrm{sin}\:{t} \\ $$$$\mathcal{G}{eneral}\:{solution}\: \\ $$$${y}\:=\:{C}_{\mathrm{1}} \mathrm{cos}\:{t}\:+\:{C}_{\mathrm{2}} \mathrm{sin}\:{t}\:+\:\mathrm{2}{t}\:\mathrm{sin}\:{t}\: \\ $$$${y}\:=\:{C}_{\mathrm{1}} \mathrm{cos}\:\left(\mathrm{ln}\:\left({x}+\mathrm{1}\right)\right)+{C}_{\mathrm{2}} \mathrm{sin}\:\left(\mathrm{ln}\:\left({x}+\mathrm{1}\right)\right)+\mathrm{2ln}\:\left({x}+\mathrm{1}\right)\mathrm{sin}\:\left(\mathrm{ln}\:\left({x}+\mathrm{1}\right)\right) \\ $$