Question

Consider the following partial differential equation $u(x,y)$ with the constant $c>1$;
$\frac{\partial u}{\partial y}+c\frac{\partial u}{\partial x}=0$
Solution of this equation is

• A. $u(x,y)=f(x+cy)$
• B. $u(x,y)=f(x-cy)$
• C. $u(x,y)=f(cx+y)$
• D. $u(x,y)=f(cx-y)$

Answer 1

The correct option is B $u(x,y)=f(x-cy)$

Let $u=f(ax+by)$
$\therefore \frac{\partial u}{\partial (ax+by)}=f^{\prime }(ax+by)$
Now, $\frac{\partial u}{\partial y}+C\frac{\partial u}{\partial x}=0$
$\Rightarrow \frac{\partial u}{\partial (ax+by)}\times \frac{\partial (ax+by)}{\partial y}+C\frac{\partial u}{\partial (ax+by)}\times \frac{\partial (ax+by)}{\partial x}=0$
$\Rightarrow b+c\times a=0$
$\Rightarrow b=-ac$
If $a=1$
$b=-c$
$\therefore u=f(1.x-c.y$)
$=f(x-cy)$
Method II: Using hit and trial method we see that option (b) satisfies given P.D.E