Can any one please explain the second step:-
Step1: $$\frac{\partial }{\partial x}\left[(1-x^2)\frac{\partial u}{\partial x}\right]+\frac{\partial }{\partial y}\left[y^2\frac{\partial u}{\partial y}\right]=0$$
Step2: $$L.H.S.=-2x\frac{\partial u}{\partial x}+(1-x^2)\frac{\partial ^2u}{\partial x^2}+2y\frac{\partial u}{\partial y}+y^2\frac{\partial ^2u}{\partial y^2}$$
Starting with Step 1 we have:- $$\frac{\partial }{\partial x}\left[1-x^2\frac{\partial u}{\partial x}\right]+\frac{\partial }{\partial y}\left[y^2\frac{\partial u}{\partial y}\right]\\=\frac{\partial}{\partial x}(1)-\frac{\partial}{\partial x}(x^2\frac{\partial u}{\partial x})+\frac{\partial}{\partial y}(y^2\frac{\partial u}{\partial y})$$
The first term of the above expression should be $0$ as we are differentiating a constant (the number $1$) with respect to $x$. Applying the product rule for differentiation to the second and third terms, we obtain:- $$0-\frac{\partial}{\partial x}(x^2\frac{\partial u}{\partial x})+\frac{\partial}{\partial y}(y^2\frac{\partial u}{\partial y})\\=-\frac{\partial}{\partial x}(x^2)\frac{\partial u}{\partial x}-x^2\frac{\partial^2u}{dx^2}+\frac{\partial}{\partial y}(y^2)\frac{\partial u}{\partial y}+y^2\frac{\partial^2u}{dy^2}\\=-2x\frac{\partial u}{\partial x}-x^2\frac{\partial^2u}{dx^2}+2y\frac{\partial u}{\partial y}+y^2\frac{\partial^2u}{dy^2}\\=\text{LHS}$$
Are you sure about the +1 after step 2? I think, that 1 is disappearing after derivation.
If you derive $-x^2\frac{\partial U}{\partial x}$ then you have to use the product rule.
$$u=-x^2$$ $$v=\frac{\partial U}{\partial x}$$ $$(u \cdot v)'=u'\cdot v+u\cdot v'$$
$$=-\frac{\partial}{\partial x}(x^2)\frac{\partial u}{\partial x}-x^2\frac{\partial^2u}{dx^2}$$
– Abdullah Sorathia Apr 22 '14 at 18:34