Solve:
$$ x^2 (a - bx) {dy^2 \over dx^2} - x (5a - 4bx) {dy \over dx} + 3(2a - bx)y = 6a^2$$
Solution (according to answer sheet):
$$ y(a-bx) = Ax^2 + Bx^3 + C$$

Looking for right substitution to convert it into nice form:
$$ u^2 {dv^2 \over du^2} + P_1 u {dv \over du} + P_2 v = F(u)$$

Question

Solve: 
$$ x^2 (a - bx) {dy^2 \over dx^2}  - x (5a - 4bx) {dy \over dx} + 3(2a - bx)y = 6a^2$$
Solution (according to answer sheet):
$$ y(a-bx) = Ax^2 + Bx^3 + C$$

Looking for right substitution to convert it into nice form:
$$ u^2 {dv^2 \over du^2}  + P_1 u {dv \over du} + P_2 v = F(u)$$

experimentx · Answer

also more thread on this Q http://math.stackexchange.com/questions/176220/need-hints-to-solve-this-de

anonymous · Answer

for the last equation ... suppose that  $u=x$  and  $v=y(a-bx)$  , $P_1=-4$ , $P_2=6$ 
and $F(u)=constant$

anonymous · Answer

then ur equation will be $$x^2 \frac{d^2v}{dx^2}-4x \frac{dv}{dx}+6v=C $$
Using Euler method answer for this equation would be 

$$v=Ax^2+Bx^3+C $$

anonymous · Answer

i tried to sub   $v=y(a-bx)$  in the original equation but it does'nt work....:(

experimentx · Answer

let me verify the solution using Maple or Mathematica ...

anonymous · Answer

ok...:)

anonymous · Answer

Let $ a=b=1$ in the ODE, then

$$
(1-x) x^2 y''(x)-(5-4 x) x
   y'(x)+3 (2-x) y(x)=6
$$

If we take
$$
y(x) =\frac{A x^2+B x^3+C}{1-x}
$$

and we replace in the original ODE and multiply the two sides by
$(1-x)^2$, we get by Mathematica
$$x^4 (7 B-A)+x^3 (4 A-3 B)+x^2
   (9 C-2 A)-3 B x^5-14 C x+6
   C=6 (x-1)^2$$
We get to solve
$$
4 A-3 B=0\
 -3 B=0\  
   7B-A=0\
    12-14C=0\
     6 C -6=0\
      -2 A+9C -C =0
$$
There is no solution for the system above.

anonymous · Answer

yeah i think there is something wrong with that...

experimentx · Answer

i guess so ... the problem in book might be wrong. I'm also getting strange answer using Maple.