solution of variable coefficient equation

Question

Consider the equation $$u_x + yu_y = 0$$ and I know that this PDE has solution $u(x,y) = f(e^{-x}y)$

Can someone help me to derive this PDE to get the solution? Thank you

Do you need to find the solution or want to prove the solution? (I did the easier of the two scenarios below) — Chinny84, Aug 14 '15 at 16:48
Make the change of variables $z=x$ and $w=e^{-x}y$. Use the chain rule. — mickep, Aug 14 '15 at 16:55
Now we have cleared up that mystery..Have you worked with method of charteristics? — Chinny84, Aug 14 '15 at 17:03

score 1 · Answer 1 · answered Aug 14 '15 at 17:20

1

Let $z=x$ and $w=e^{-x}y$. This is a change of coordinates, since the jacobian is $e^{-x}\neq 0$. Also, let $v(z,w)=u(x,y)$. Then, by the chain rule, $$ u'_x=v'_z-ye^{-x}v'_w\quad\text{and}\quad u'_y=e^{-x}v'_w. $$ Hence, $$ 0=u'_x+yu'_y=v'_z, $$ so $v(z,w)$ is a function of $w$ only, say $v(z,w)=f(w)$. This means that $$ u(x,y)=v(z,w)=f(w)=f(e^{-x}y). $$

answered Aug 14 '15 at 17:20

mickep

19,962

assume that you don't know the solution is $w$. how can you let $w = e^{-x}y?$ – Alexander Aug 14 '15 at 17:28
You can use the method of characteristics to find this out. See for example this question. – mickep Aug 14 '15 at 19:10

score 1 · Accepted Answer · answered Aug 15 '15 at 16:27

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dx}{dt}=1$ , letting $x(0)=0$ , we have $x=t$

$\dfrac{dy}{dt}=y$ , letting $y(0)=y_0$ , we have $y=y_0e^t=y_0e^x$

$\dfrac{du}{ds}=0$ , letting $u(0)=f(y_0)$ , we have $u(x,y)=f(y_0)=f(e^{-x}y)$

solution of variable coefficient equation

2 Answers2