Solve the initial value problem : $u_{xy}=6x^2y$ ,$u(x;0)=1-\cos x$ , $u(0;y)=3y^2$ .
I was trying like this , there is a special solution for $u_{xy}=6x^2y$ : $u=x^3y^2$. And for $u_{xy}=0$, we can get $u(x,y)=f(x)+g(y)$. But after substitute it the initial value , there always has a constant in it. How to do it next?