Re: system of partial DE's

The notation is $\displaystyle \frac{\partial u}{\partial y}=2xyu$ etc...

Your solution idea is confusing as stated, what do you mean by take the derivative of the first one? Derivative with respect to what? You mean take $\displaystyle \frac{\partial}{\partial x}\frac{\partial u}{\partial y}=\frac{\partial}{\partial y}\frac{\partial u}{\partial x}$ ?

Re: system of partial DE's

Actually I tried it and it does noting, lol, it just takes you back to where you started but it would be interesting if someone knew what to do with this.

Re: system of partial DE's

In $\displaystyle \frac{\partial u}{\partial x}= (y^2+ 5)u$, y can be treated as a constant: think of it as $\displaystyle \frac{du}{dx}= (y^2+ 5)u$ and **separate**: $\displaystyle \frac{du}{u}= (y^2+ 5)dx$ where, remember y is a constant. Integrate both sides to get $\displaystyle ln|u|= (y^2+ 5)x+ f(y)$. (Since "y is a constant", the constant of integration may be function of y, f(y).)

We can solve that for u as $\displaystyle u= F(y)e^{(y^2+ 5)x}$ where $\displaystyle F(y)= e^{f(y)}$. Since f(y) is an arbitrary function of y, so is F(y).

With that we can write $\displaystyle \frac{\partial u}{\partial y}= F'(y)e^{(y^2+ 5)x}+ F(y)(2xy)e^{(y^2+ 5x)}= 2xyu= 2xye^{(y^2+ 5)x}$ which, since F cannot depend on x, requires that F(y)= 1 (so that F'(y)= 0).

Re: system of partial DE's

That simplifies things, I was not treating y as a constant.