$(x+1)^2y''+(x+1)y'+y=x^2+2\sin(\ln(x+1)), y(0)=\frac{1}{5},y'(0)=2$

$(x+1)^2y''+(x+1)y'+y=x^2+2\sin(\ln(x+1)), y(0)=\frac{1}{5},y'(0)=2$

My solution:

$y''+\frac{y'}{(x+1)}+\frac{y}{(x+1)^2}=\frac{x^2+2\sin(\ln(x+1))}{(x+1)^2}$

First of all , I found the equation solution of $y''+\frac{y'}{(x+1)}+\frac{y}{(x+1)^2}=0$

$y=c_1\cos(\ln(x+1))+c_2\sin(\ln(x+1))$

I try to solve this ode using the variation of parameters theorem

Get this system equation:

(I)$c'_1\cos(\ln(x+1))+c'_2\sin(\ln(x+1))=0$

(II)$-c'_1\sin(\ln(x+1))+c'_2\cos(\ln(x+1))=x^2+2\sin(\ln(x+1))$

Multiply (I) by $\sin(\ln(x+1))$ , (II) by $\frac{\cos(\ln(x+1))}{x+1}$.

By addtion i get:

$c'_2=\frac{\cos(\ln(x+1))[x^2+2\sin(\ln(x+1))]}{(x+1)}$

I do not know how I get $c_2$ by an integral ?

Help please

Thanks !

The two equations you have using variation of parameters is incorrect. Given a second-order linear inhomogeneous DE $y'' + p(x)y' + q(x)y = f(x)$ with homogeneous solution $y_c(x) = c_1y_1(x) + c_2y_2(x)$, a particular solution $y_p$ is given by $y_p(x) = u_1(x)y_1(x) + u_2(x)y_2(x)$, where $u_1$ and $u_2$ satisfy \begin{align*} u_1'y_1 + u_2'y_2 & = 0 \\ u_1'y_1' + u_2'y_2' & = f(x). \end{align*} Let $y_1 = \cos\left(\ln(x + 1)\right)$ and $y_2 = \sin\left(\ln(x + 1)\right)$. We have that \begin{align} \tag{1} u_1'\cos\left(\ln(x + 1)\right) + u_2'\sin\left(\ln(x + 1)\right) & = 0 \\ \tag{2} -u_1'\frac{\sin\left(\ln(x + 1)\right)}{x + 1} + u_2'\frac{\cos\left(\ln(x + 1)\right)}{x + 1} & = \frac{x^2 + 2\sin\left(\ln(x + 1)\right)}{(x + 1)^2} \end{align} Multiplying $(1)$ by $\sin\left(\ln(x + 1)\right)$, $(2)$ by $(x + 1)\cos\left(\ln(x + 1)\right)$, and adding the two resulting equations, we get \begin{align*} u_2' & = \frac{x^2 + 2\sin\left(\ln(x + 1)\right)}{x + 1}\cdot \cos\left(\ln(x + 1)\right). \end{align*} Using the substitution $z = \ln(x + 1)$, we see that $$ u_2 = \int \left[\left(e^z - 1\right)^2 + 2\sin z\right]\cos z\, dz. $$ Computing this integral is a standard exercise now. Can you take it from here?

$$(x+1)^2y''+(x+1)y'+y=x^2+2\sin(\ln(x+1))$$ $$y(0)=\frac{1}{5},y'(0)=2$$ Substitute $u=x+1$: $$u^2y''+uy'+y=(u-1)^2+2\sin(\ln(u))$$ $$y(1)=\frac{1}{5},y'(1)=2$$ Substitute $u=e^t$: $$y''(t)+y(y)=(e^t-1)^2+2\sin(t)$$ This is easy to solve with variation of constants method. $$y''(t)+y(t)=2 \sin t$$ $$\implies y_p=At\cos t$$ $$y''(t)+y(t)=e^{2t}-2e^t+1$$ $$y_p=Ae^{2t}+Be^t+C$$

Edit:

Note that: $$\dfrac {dy}{du}=\dfrac {dy}{dt}\dfrac{dt}{du}=e^{-t} \dfrac {dy}{dt}=\dfrac 1u\dfrac{dy}{dt}$$

Then $$u\dfrac {dy}{du}=\dfrac {dy}{dt}$$ Do the same for: $$\dfrac {d^2y}{du^2}$$ Then the DE is simplified and you find that : $$y''(t)+y(t)=(e^t-1)^2+2\sin(t)$$