Solve $x^2 (a - bx) {d^2y \over dx^2} - x (5a - 4bx) {dy \over dx} + 3(2a - bx)y = 6a^2$ What do I substitute here
$$ x^2 (a - bx) {d^2y \over dx^2}  - x (5a - 4bx) {dy \over dx} + 3(2a - bx)y = 6a^2 $$ to get it into this form?
$$ u^2 {dv^2 \over du^2}  + P_1 u {dv \over du} + P_2 v = F(u) $$
The solution (according to answer sheet):
$y(a-bx) = Ax^2 + Bx^3 + C$
 A: I will post a partial answer, as I might have made a sign error or been taken astray at some point. The method amounts to finding some sort of integrating factor for the equation. It seems reasonable to start with the change of the dependent variable:
$$u=a-bx$$
$$x=-\frac{1}{b}\left(u-a\right)$$
$$\frac{dy}{dx}=-b\frac{dy}{du}$$
$$\frac{d^{2}y}{dx^{2}}=-b^{2}\frac{d^{2}y}{du^{2}}$$
Rewrite the equation:
$$\left(u-a\right)^{2}u\frac{d^{2}y}{du}+\left(u-a\right)\left(4u+a\right)\frac{dy}{du}+3\left(u+a\right)y=6a^{3}$$
Split the $\frac{dy}{du}$ term and regroup:
$$\left(u-a\right)^{2}u\frac{d^{2}y}{du}+\left(u-a\right)\left(u+a\right)\frac{dy}{du}+3\left(u-a\right)u\frac{dy}{du}+3\left(u+a\right)y=6a^{3}$$
$$\left(u-a\right)\left[\left(u-a\right)u\frac{d^{2}y}{du^{2}}+\left(u+a\right)\frac{dy}{du}\right]+3\left[\left(u-a\right)u\frac{dy}{du}+\left(u+a\right)y\right]=6a^{3}$$
Differentiating the second square bracket:
$$\frac{d}{du}\left[\left(u-a\right)u\frac{dy}{du}+\left(u+a\right)y\right]=\left(2u-a\right)\frac{dy}{du}+\left(u-a\right)u\frac{d^{2}y}{du^{2}}+y+\left(u+a\right)y\qquad(1)$$
We can express the first one as follows:
$$\left(u-a\right)u\frac{d^{2}y}{du^{2}}+\left(u+a\right)\frac{dy}{du}=\frac{d}{du}\left[\left(u-a\right)u\frac{dy}{du}+\left(u+a\right)y\right]-\left(2u-a\right)\frac{dy}{du}-y$$
Inserting in the equation:
$$\left(u-a\right)\frac{d}{du}\left[\left(u-a\right)u\frac{dy}{du}+\left(u+a\right)y\right]-\left(u-a\right)\left[\left(2u-a\right)\frac{dy}{du}+y\right]+3\left[\left(u-a\right)u\frac{dy}{du}+\left(u+a\right)y\right]=6a^{3}$$
$$\left(u-a\right)\frac{d}{du}\left[\left(u-a\right)u\frac{dy}{du}+\left(u+a\right)y\right]+\left(u^{2}-a^{2}\right)\frac{dy}{du}+2ay=6a^{3}$$
Now using the product rule we obtain:
$$\left(u-a\right)\frac{d}{du}\left[\left(u-a\right)u\frac{dy}
{du}+\left(u+a\right)y\right]=\frac{d}{du}\left[\left(u-a\right)^{2}u\frac{dy}{du}+\left(u^{2}-a^{2}\right)y\right]-\left(u-a\right)u\frac{dy}{du}-\left(u+a\right)y$$
The original equation is then expressed as follows:
$$\frac{d}{du}\left[\left(u-a\right)^{2}u\frac{dy}{du}+\left(u^{2}-a^{2}\right)y\right]+\left(u-a\right)\left[a\frac{dy}{du}-y\right]=6a^{3}$$
Now it might have made sense in (1) to look for $y$ in some specific form, say $y=rs$ and then impose suitable conditions on the factors.
