2nd order linear differential equation

**acevipa** · September 17th 2010, 06:50 PM

Can someone check my working for this question

By making the substitution $x=e^t$ , reduce the following linear differential equation to one with constant coefficients, and hence solve it:

$x^2\frac{d^2y}{dx^2}-4x\frac{dy}{dx}+6y=x^5 \ \ \ \ \ \ , x>0$

This is what I've done so far:

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

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

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

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

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

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

$e^{2t}\left(e^{-2t}\,\frac{d^2y}{dt^2} - e^{-2t}\,\frac{dy}{dt}\right)-4e^t\left(e^{-t}\frac{dy}{dt}\right)+6y=e^{5t}$

$\frac{d^2y}{dt^2}-5\frac{dy}{dt}+6y=e^{5t}$

Then would you solve that like a normal ODE

**Prove It** · September 17th 2010, 06:53 PM

Everything looks fine

BTW thank you for making an effort to do the problem yourself and showing all your working

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