Solution differential equation via Laplace transform

Question

Please help me finish this problem.

$xy''+(3x-1)y'-(4x+9)y=0$ where $y(0)=0$

$L[xy'']+L[(3x-1)y']-L[(4x+9)y]=L[0]$

$L[xy'']=\frac{d}{dp}(p^2Y)$

$L[(3x-1)y']=-3\frac{d}{dp}(pY)$

$L[(4x+9)y]=-4\frac{dY}{dp}$

$-\frac{d}{dp}(p^2Y)-3\frac{d}{dp}(pY)+4\frac{dY}{dp}=0$

$-\frac{d}{dp}(p^2Y)-(3pY)\frac{d}{dp}+4\frac{dY}{dp}=0$

I don't know what to do from here and would really appreciate some help finishing the problem.

Answer 1

Hint:

$$\int \frac{dY}{3Y} = \int\frac{dp}{4-p^2}.$$

The first integral is trivial, you should be okay from here. The second can be equivalently expressed as

$$ \int\frac{dp}{4-p^2} \equiv \frac{1}{4}\int \left( \frac{1}{(2+p)} - \frac{1}{(2-p)} \right) dp.$$

You should be able to take it from here.

Answer 2

Taking the Laplace transform of the differential equation, you should get $$ \left( -{p}^{2}-3\,p+4 \right) {\frac {\rm dY}{{\rm d}p}} -3\,pY -12\,Y = 0 $$ and factoring out $p+4$, $$ (p-1) \dfrac{dY}{dp} + 3 Y = 0$$ That's a separable first-order equation, which you should be able to solve...