# General solution for third order differential equation

I am having trouble starting the following problem for the general solution:

$x^3y''' + x^2y'' -2xy' + 2y = 2x^4$ for $x>0$

I feel like this can be simplified to an equation that is easier to work with, but I'm not sure what that change of variable or simplification is. Any hints or suggestions?

Thanks

The homogeneous differential equation $x^3 y''' + x^2 y'' - 2 x y' + 2 y = 0$ is a third order Cauchy-Euler differential equation. The thing to do here is to look for solutions of the form $y = x^p$. You will find three such $p$. Then, since $x^4$ is not a solution of the homogeneous equation, your inhomogeneous equation should have a particular solution of the form $y = C x^4$ for some constant $C$. Add this particular solution to the general solution of the homogeneous equation and you have the general solution of your equation.
You are right that there is an easy way of solving this. The key to many differential equations is to notice the product rules. Allow me to show you. $$x^3y''' + x^2y'' -2xy' + 2y = 2x^4$$ $$x^3y'''+3x^2y''- 2x^2y'' - 4xy' + 2xy' + 2y = 2x^4$$ $$(x^3y'')' - 2(x^2y')' + 2(xy)' = 2x^4$$ Now integrating gives us a second order differential equation! $$x^3y'' - x^2y' + 2xy=\frac{2}{5}x^5+C_1$$ I'll leave you to attacking this one, as it is only a second-order ODE. Let me know if you need some more help!