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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.

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  • $\begingroup$ Try $e^x(ax^4 + bx^3 + cx^2 + dx + e)$ $\endgroup$ – Frederick Feb 7 '14 at 8:30
  • $\begingroup$ Whoops, meant $2x e^x$. $\endgroup$ – Frederick Feb 7 '14 at 8:48
  • $\begingroup$ Following Frederick's track, $y=2 e^x x + c_ 1 x + c_ 2 x^2 + c_ 3 x^3$ $\endgroup$ – Claude Leibovici Feb 7 '14 at 8:52
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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 $$ z(x)=c_1\mathrm{e}^x+c_2\mathrm{e}^{2x}+c_3\mathrm{e}^{3x}. $$

Then variation of parameters and inverse transformation.

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