Solve the differential equation $(x+1)^2y''-2(x+1)y'+2y=0$ Solve the differential equation $$(x+1)^2y''-2(x+1)y'+2y=0$$
My attemp:
sub. $x+1=u$
after this substitution the above equation takes form
$$u^2y''-2uy'+2y=0$$
This equation is of the form $ax^2y''+bxy'+cy=0$-which solution is $ar^2+(b-a)r+c=0.$ According to this for our equation we have $a=1,b=-2,c=2$
Now i have
$$ar^2-2r+2=0$$
Solution of end equation is $r=1\pm i,$  where $\alpha =1, \beta=1$
Solution is:
$$y_{H}=C_1u\cos(\ln x)+C_2u\sin(\ln x)=C_1(x+1)\cos(\ln x)+C_2(x+1)\sin(\ln x)$$
But i didnt now its the correct answer.Tell me please.
 A: $$(x+1)^2y''-2(x+1)y'+2y=0$$
Rewrite it as;
$$y''-2\dfrac {(x+1)y'-y}{(x+1)^2}=0$$
$$y''-2\left (\dfrac {y}{x+1}\right)'=0$$
And integrate.
A: Your solution is wrong, $u^2y''-2uy'+2y=0$ is not an ODE with constant coefficients, so you cannot use the characteristic equation directly like this.
Fortunately in the case of $u^py^{(p)}$ groups where $y^{(p)}$ denotes the $p-$derivarive, and $x^p$ powers of $x$ with the same $p$ then the substitution $y=f(\ln(u))$ makes things pretty.
$\begin{cases}y=f(\ln(u))\\
y'=\frac 1uf'(\ln(u))\\
y''=\frac {-1}{u^2}f'(\ln(u))+\frac 1{u^2}f''(\ln(u))\end{cases}$
Let set $t=\ln(u)$ and then $\Big(-f'(t)+f''(t)\Big)-2f'(t)+2f(t)=0$
Your ODE is transformed to a linear one with constant coefficients
$$f''(t)-3f'(t)+2f(t)=0$$
Characteristic equation is $r^2-3r+2=(r-1)(r-2)$ and we have $f(t)=Ae^t+Be^{2t}$
Substituting back $t=\ln(u)$ and $u=x+1$ you get
$$y(x)=A(x+1)^2+B(x+1)$$
A: Hint: define $z:=y/u$ so$$y=uz\implies y^\prime=uz^\prime+z,\,y^{\prime\prime}=uz^{\prime\prime}+2z^\prime\implies0=u^2y^{\prime\prime}-2uy^\prime+2y=u^3z^{\prime\prime}.$$
