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When we have the non-homogeneous differential equation $$ay''(x)+by'(x)+cy(x)=f(x)$$ and the non-homogeneous term $f(x)$ is of the form $d_1e^{mx}$ we know that the particular solution is $$y_p=d_2x^ke^{mx}$$ where $k$ is the multiplicity of the eigenvalue $w=m$.

When we have the non-homogeneous differential equation $$ay''(x)+by'(x)+cy(x)=\sum_{i=1}^n s_i e^{r_ix}$$ will the particular solution be $$y_p=x^k\sum_{i=1}^n h_ie^{r_ix}$$ ?

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    $\begingroup$ They are called particular solutions, not partial solutions. $\endgroup$ – Julián Aguirre Oct 27 '15 at 11:16
  • $\begingroup$ Oh sorry, I will edit it... @JuliánAguirre $\endgroup$ – Mary Star Oct 27 '15 at 11:25
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The particular solution will be of the form $$ y_p=\sum_{i=1}^nh_i\,x^{k_i}e^{r_ix} $$ where $k_i$ is the multiplicity of $r_i$.

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  • $\begingroup$ The characteristic equation of the corresponding homogeneous problem is $$a\lambda^2+b\lambda+c=0$$ The discriminant is $$\Delta=b^2-4ac$$ and then we have $$\lambda_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$ So that the solution contains only exponentials, we want that $b^2-4ac \neq 0$, right? That means that $k_i=0, \forall i$, or not? $\endgroup$ – Mary Star Oct 27 '15 at 11:40
  • $\begingroup$ Or is what I said wrong? $\endgroup$ – Mary Star Oct 27 '15 at 15:49
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    $\begingroup$ The particular solution contains only exponentials because the right hand side of the equation contains only exponentials. The solution of the homogeneous equation may contain sines and cosines. $\endgroup$ – Julián Aguirre Oct 27 '15 at 17:04
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    $\begingroup$ The multiplicity of each root of the characteristic equation (what you call eigenvalue) is $1$ in that case. If $r_i$ is complex and is a solution of the characteristic equation, you will have terms of the form $x\,e^{r_ix}$. $\endgroup$ – Julián Aguirre Oct 27 '15 at 17:18
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    $\begingroup$ That is correct. $\endgroup$ – Julián Aguirre Oct 27 '15 at 21:13

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