# How to solve the solution to this third order nonhomogeneous ODE

I am supposed to solve this nonhomogeneous ordinary differential equation.

$$x^3\frac{d^3y}{dx^3}-3x^2\frac{d^2y}{dx^2}+6x\frac{dy}{dx}-6y=2x^4e^x$$

I attempted in solving this problem through finding the particular and the homogeneous solution. In finding the particular solution, I have attempted in using the method of undetermined coefficients. However, I was not sure of what function to use in finding the particular solution.

• Try $e^x(ax^4 + bx^3 + cx^2 + dx + e)$ – Frederick Feb 7 '14 at 8:30
• Whoops, meant $2x e^x$. – Frederick Feb 7 '14 at 8:48
• Following Frederick's track, $y=2 e^x x + c_ 1 x + c_ 2 x^2 + c_ 3 x^3$ – Claude Leibovici Feb 7 '14 at 8:52

Hint. Transformation $y(x)=z(\log x)$. Then $$y'=\frac{z'(\log x)}{x},\quad y''=\frac{z''(\log x)-z'(\log x))}{x^2},\quad y'''=\frac{z'''(\log x)-3z''(\log x)+2z'(\log x)}{x^3}.$$ Then our equation becomes: \begin{align} x^3\frac{d^3y}{dx^3}-3x^2\frac{d^2y}{dx^2}+6x\frac{dy}{dx}-6y &= \left(z'''-3z''+2z'\right)- 3\left(z''-z'\right)+6z'-6z \\ &= z'''-6z''+11z'-6z, \end{align} and it is finally transformed as $$z'''(\log x)-6z''(\log x)+11z'(\log x)-6z(\log x)=x^3\mathrm{e}^x,$$ or $$z'''(x)-6z''(x)+11z'(x)-6z(x)=\mathrm{e}^{3x}\mathrm{e}^{\mathrm{e}^x}.$$ The general solution of the homogeous equation is $$z(x)=c_1\mathrm{e}^x+c_2\mathrm{e}^{2x}+c_3\mathrm{e}^{3x}.$$