# solving (x+a)^2 d2y/dx2 -4(x+a) dy/dx + 6y = x

#### sapinder

Hi, my question is that can we find the particular integral of this kind of equation by variation of parameters? If yes, then what mistake did I make in my solution in the provided link? If this is not the method to adopt, what other method can we use?

#### idontknow

$$\displaystyle (x+a)^2 y'' -4(x+a) y' + 6y = x$$.
Set $$\displaystyle x+a=e^{u}$$.

#### idontknow

H-homogeneous solution , p-particular solution .
$$\displaystyle Hâ€™â€™-5Hâ€™+6H=0$$ ; $$\displaystyle H=c_1 e^{2u } +c_2 e^{3u}.$$

$$\displaystyle \begin{cases}p=v_1 (u)e^{2u} +v_2 (u) e^{3u} \\ pâ€™â€™-5pâ€™+6p-x=0 \end{cases}$$

Solve the system of equations for $$\displaystyle v_1 , v_2$$.
The general solution is $$\displaystyle y=H+p$$.