How I can calculate this partial derivative of $f(a,b)=\int_0^\infty  e^{-ax^3-bx^2}\mathrm dx$? My question is: 
How to prove that the function:
$$f(a,b)=\int_0^\infty  e^{-ax^3-bx^2}\mathrm dx$$
is a solution of the differential equation:
$$3ab\frac{{{\partial ^2}f}}{{\partial {b^2}}} - 3a\frac{{\partial f}}{{\partial b}} - 2{b^2}\frac{{\partial f}}{{\partial a}} = 1 ?$$
I have reviewed in several sites, but still nothing. :(
Thanks in advance :)
 A: If the goal of the question is to prove that 
$$f(a,b) = \int_0^\infty e^{-ax^3-bx^2}\mathrm dx,$$
satisfies
$$3ab\frac{\partial^2 f}{\partial b^2}-3a\frac{\partial f}{\partial b}-2b^2\frac{\partial f}{\partial a} = 1,$$
then this can be done by looking at all terms at the same time, not separately.
I will assume that the differentiating under the integral sign is not the problematic part.
By plugging in, we get the following:
$$3ab\frac{\partial^2f}{\partial b^2}-3a\frac{\partial f}{\partial b}-2b^2\frac{\partial f}{\partial a} = \int_0^\infty \left(3abx^4+3ax^2+2b^2 x^3\right)e^{-ax^3-bx^2}\mathrm dx.$$
Now, observe that 
$$\int_0^\infty 3ax^2e^{-ax^3 - bx^2}\mathrm dx = \int_0^\infty ax^3\left(3ax^2 + 2bx\right)e^{-ax^3-bx^2}\mathrm dx ,$$
by integration by parts.
This means that, after factoring, we obtain
$$\int_0^\infty \left(3abx^4+3ax^2+2b^2 x^3\right)e^{-ax^3-bx^2}\mathrm dx = \int_0^\infty \left(3ax^2+2bx\right) \left(ax^3+bx^2\right) e^{-ax^3-bx^2}\mathrm dx.$$
Using the substitution $u = ax^3 + bx^2$, this reduces to
$$\int_0^\infty ue^{-u}\mathrm du = 1,$$
whence we get the wanted identity.
