Given $z = u(x, y)e^{ax+by}$ and $\frac{\partial^2u}{\partial x\partial y}=0$. Find values of the constants $a$ and $b$ such that ...

Question

This problem is from Apostol Calculus Vol.2, page 255, exercise 19.

Given $z = u(x, y)e^{ax+by}$ and $\frac{\partial^2u}{\partial x\partial y}=0$. Find values of the constants $a$ and $b$ such that $\frac{\partial^2z}{\partial x\partial y}-\frac{\partial z}{\partial x}-\frac{\partial z}{\partial y}+z=0$

I'm still fairly new to multivariable calculus and diferential equations, but so far what I've done is partially derived the three formulas and replaced them in the original equation, what I got was this:

$(b-1)\frac{\partial u}{\partial x}-\frac{\partial u}{\partial y}=-(ab-a-b+1)u$

Don't know how to procced from there, any help would be appreciate it.

Answer 1

You're on the right track. Seems like you have made a small error:

\begin{align} z & = ue^{ax+by} \\ \frac{\partial z}{\partial x} & = \bigg(au+\frac{\partial u}{\partial x}\bigg)e^{ax+by} \\ \frac{\partial z}{\partial y} & = \bigg(bu+\frac{\partial u}{\partial y}\bigg)e^{ax+by} \\ \frac{\partial^2z}{\partial x \partial y} & = \bigg(abu+a\frac{\partial u}{\partial y} + b\frac{\partial u}{\partial x}+ \frac{\partial^2u}{\partial x \partial y} \bigg)e^{ax+by} \end{align}

So if you plug it into the equation, you should get

$$(b-1)\frac{\partial u}{\partial x}+(a-1)\frac{\partial u}{\partial y} = -(ab-a-b-1)u$$

The RHS can be factorised, giving

$$(b-1)\frac{\partial u}{\partial x}+(a-1)\frac{\partial u}{\partial y} = -(a-1)(b-1)u$$

Note that the question asks for this to be true for ALL $u(x,y)$. So in fact, we must require

$$b-1=a-1=(a-1)(b-1)=0$$

Thankfully, this is possible if $a=b=1$.

Answer 2

Something went wrong with your computations. I got:

$$ (*) \quad (b-1)\frac{\partial u}{\partial x}+(a-1)\frac{\partial u}{\partial y}+(ab-a-b+1)u=0.$$

Hence $(*)$ holds $ \iff a=b=1.$