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Can you please help me to solve this question?

Find 2 solutions of equation:

$\dfrac{\partial u}{\partial x}\dfrac{\partial u}{\partial y}=1$


Thank you.

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2 Answers 2

up vote 2 down vote accepted

In fact this belongs to a PDE of the form http://eqworld.ipmnet.ru/en/solutions/fpde/fpde3308.pdf.

The complete integral (trivial solution) is $u=ax+\dfrac{y}{a}+b$

According to http://books.google.com.hk/books?id=hkWDQ57NlksC&pg=PA1&dq=Partial+Differential+Equations+by+Bhamra&hl=zh-CN&sa=X&ei=8mhPUZeYBciaiAe3m4HADw&ved=0CDEQ6AEwAA in page $90$, the general integral (general solution) is $\begin{cases}u=ax+f(a)y+g(a)\\x+\dfrac{\partial f(a)}{\partial a}y+\dfrac{\partial g(a)}{\partial a}=0\end{cases}$

$u(2x,2x)=5x$ :

$\begin{cases}2ax+2f(a)x+g(a)=5x\\2x+2\dfrac{\partial f(a)}{\partial a}x+\dfrac{\partial g(a)}{\partial a}=0\end{cases}$

$\begin{cases}(2a-5+2f(a))x=-g(a)\\\biggl(2+2\dfrac{\partial f(a)}{\partial a}\biggr)x=-\dfrac{\partial g(a)}{\partial a}\end{cases}$

$\therefore\begin{cases}u=ax+f(a)y+g(a)\\x+\dfrac{\partial f(a)}{\partial a}y+\dfrac{\partial g(a)}{\partial a}=0\end{cases}$ , where $\begin{cases}(2a-5+2f(a))x=-g(a)\\\biggl(2+2\dfrac{\partial f(a)}{\partial a}\biggr)x=-\dfrac{\partial g(a)}{\partial a}\end{cases}$

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HINT: Assume $u(x,y)$ is a function of form $$u(x,y)=ax+by+c$$ From $u(2x,2x)=5x$, we have that $u(0,0)=0$ and subsequently $c=0$. Hence $$u(x,y)=ax+bx$$ Now use the two equations you have to find $a$ and $b$.

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