Find A Particular Integral

Question

I'm trying to find the particular integral for the equation:

$$P(D)y=Q(x)=x+e^x$$

My initial thought was to use $(PI)(x)=y(x)=ax+bxe^x$ but it doesn't seem to work. Any help would be greatly appreciated, thank you very much!

Here's the full question:

Answer 1

• $x$: try a solution of the form $y=x$. Then $y'''-3y''+9y'+13y$ yields $13x+9$, and you can adjust by deducting $9/13$;

• $e^x$: try a solution of the form $y=e^x$. Then $y'''-3y''+9y'+13y$ yields $20e^x$.

Hence

$$y=\frac{x}{13}-\frac9{169}+\frac{e^x}{20}.$$

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When plugged in the LHS, a polynomial gives a polynomial of the same degree.

When plugged in the LHS, an exponential gives an exponential of the same coefficient, or zero (when the coefficient is a root of the polynomial).

Only in case of a zero you need to multiply by a polynomial of a degree equal to the multiplicity of the root.

Answer 2

Here is the solution below and contains the Particular Integral.