1
$\begingroup$

I know this isn't supposed to be difficult, but I'm not sure how to use the chain rule.

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be differentiable and define $u: \mathbb{R}^2 \rightarrow \mathbb{R}$ by $u(x,y):=e^{2x}f(ye^{-x})$. I want to show that $u$ satisfies the partial differential equation $$\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y}=2u.$$

I know how to take partial derivatives with respect to $x$ and $y$, but I don't know how to implement the chain rule here. How do I do this? I appreciate your help.

$\endgroup$
3
  • 1
    $\begingroup$ Do you know how to use the chain rule to find $(g\circ h)'$? $\endgroup$
    – Git Gud
    May 31 '13 at 19:24
  • 1
    $\begingroup$ $(g\circ h)'(x)=g'(h(x))h'(x)$ $\endgroup$
    – user4167
    May 31 '13 at 19:29
  • $\begingroup$ Prove the following equation via partial differentiations of $u(x,y)$ by Git Gud's hint. $\endgroup$
    – NasuSama
    May 31 '13 at 19:30
1
$\begingroup$

Here is how one starts off proving the equality.

Consider differentiating the function with respect to certain variable... We have:

$$\frac{\partial u}{\partial x} = 2e^{2x}f(ye^{-x}) + (-ye^{-x}e^{2x}f'(ye^{-x}))$$ $$= 2e^{2x}f(ye^{-x}) - ye^{x}f'(ye^{-x})$$ $$\frac{\partial u}{\partial y} = e^{-x}e^{2x}f'(ye^{-x})$$

Substitute the given and the following equations for the given equality, which yields:

$$LHS = 2e^{2x}f(ye^{-x}) - ye^{x}f'(ye^{-x}) + y(e^{-x}e^{2x}f'(ye^{-x}))$$ $$= 2e^{2x}f(ye^{-x}) - ye^{x}f'(ye^{-x}) + ye^{x}f'(ye^{-x})$$ $$= 2e^{2x}f(ye^{-x})$$ $$= 2(e^{2x}f(ye^{-x}))$$ $$= 2u = RHS$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.