# Solving differential equation

• Sep 4th 2010, 01:10 AM
acevipa
Solving differential equation
Solve $\displaystyle x^2y''+xy'+y=\log x \ \ \ \ x>0$ by letting $\displaystyle x=e^t$

How do I find y' and y''?

$\displaystyle \frac{dy}{dx}=\frac{dy}{dt}\cdot\frac{dt}{dx}$

$\displaystyle \frac{dy}{dx}=\frac{dy}{dt}\cdot e^{-t}$

Not too sure what to do next?
• Sep 4th 2010, 01:20 AM
Prove It
Your calculation of $\displaystyle \frac{dy}{dx}$ is correct.

So $\displaystyle \frac{d^2y}{dx^2} = \frac{d}{dx}\left(e^{-t}\,\frac{dy}{dt}\right)$

$\displaystyle = e^{-t}\,\frac{d}{dx}\left(\frac{dy}{dt}\right) + \frac{d}{dx}(e^{-t})\,\frac{dy}{dt}$

$\displaystyle = e^{-t}\,\frac{d}{dt}\left(\frac{dy}{dt}\right)\,\frac{ dt}{dx} + \frac{d}{dt}(e^{-t})\,\frac{dt}{dx}\,\frac{dy}{dt}$

$\displaystyle = e^{-2t}\,\frac{d^2y}{dt^2} - e^{-2t}\,\frac{dy}{dt}$.

Now substitute everything into your DE.
• Sep 4th 2010, 01:23 AM
acevipa
Quote:

Originally Posted by Prove It
Your calculation of $\displaystyle \frac{dy}{dx}$ is correct.

So $\displaystyle \frac{d^2y}{dx^2} = \frac{d}{dx}\left(e^{-t}\,\frac{dy}{dt}\right)$

$\displaystyle = e^{-t}\,\frac{d}{dx}\left(\frac{dy}{dt}\right) + \frac{d}{dx}(e^{-t})\,\frac{dy}{dt}$

$\displaystyle = e^{-t}\,\frac{d}{dt}\left(\frac{dy}{dt}\right)\,\frac{ dt}{dx} + \frac{d}{dt}(e^{-t})\,\frac{dt}{dx}\,\frac{dy}{dt}$

$\displaystyle = e^{-2t}\,\frac{d^2y}{dt^2} - e^{-2t}\,\frac{dy}{dt}$.

Now substitute everything into your DE.

Thanks, that makes a lot of sense now.