# Thread: solve this differential equation

1. ## solve this differential equation

2. Originally Posted by szpengchao
well, it's all about chain rule! looking at $\displaystyle u$ as a function of $\displaystyle \xi, \ \eta,$ and each of $\displaystyle \xi$ and $\displaystyle \eta$ as functions of $\displaystyle x, \ y,$ and

applying the chain rule we'll get: $\displaystyle \frac{\partial u}{\partial x} = \frac{\partial u}{\partial \xi} + \frac{\partial u}{\partial \eta},$ and $\displaystyle \frac{\partial u}{\partial y}=2 \frac{\partial u}{\partial \xi} + 3\frac{\partial u}{\partial \eta}.$ now apply the chain rule again, but

this time for $\displaystyle \frac{\partial u}{\partial x}$ and $\displaystyle \frac{\partial u}{\partial y},$ to get: $\displaystyle \frac{\partial^2 u}{\partial x^2}=\frac{\partial^2 u}{\partial \xi^2}+2\frac{\partial^2 u}{\partial \xi \partial \eta} + \frac{\partial^2 u}{\partial \eta^2}, \ \ \ \frac{\partial^2 u}{\partial y^2}=4\frac{\partial^2 u}{\partial \xi^2}+12\frac{\partial^2 u}{\partial \xi \partial \eta} + 9\frac{\partial^2 u}{\partial \eta^2},$ and finally:

$\displaystyle \frac{\partial^2 u}{\partial x \partial y}=2\frac{\partial^2 u}{\partial \xi^2}+5\frac{\partial^2 u}{\partial \xi \partial \eta} + 3\frac{\partial^2 u}{\partial \eta^2}.$ plugging these three in the equation your problem has given us, will give you:

$\displaystyle \frac{\partial^2 u}{\partial \xi \partial \eta}=-1,$ which obviously has the general solution $\displaystyle u=-\xi \eta + f(\xi) + g(\eta),$ which, in terms of $\displaystyle x, y$ becomes:

$\displaystyle (*) \ \ \ u(x,y)=-(x+2y)(x+3y)+f(x+2y)+g(x+3y).$ for the required specific function, first see that:

$\displaystyle \frac{\partial u}{\partial y}=-(5x+12y)+2f'(x+2y)+3g'(x+3y).$ we'd like to have $\displaystyle u(x,0)=0,$ i.e. $\displaystyle f(x)+g(x)=x^2. \ \ \ \ (1)$

we also have this condition that $\displaystyle \frac{\partial u}{\partial y}(x,0)=x,$ which gives us: $\displaystyle 2f'(x)+3g'(x)=6x. \ \ \ \ (2)$

solving the system of equations (1) and (2) gives us $\displaystyle f(x)=C$ and $\displaystyle g(x)=x^2 - C,$ where $\displaystyle C$ is a constant.thus

by $\displaystyle (*)$ the specific solution is $\displaystyle u(x,y)=-(x+2y)(x+3y)+(x+3y)^2=y(x+3y). \ \ \ \square$