In general, particular solution to differential equation is any solution that fulfills the equation. For example, setting y (our dependent variable) to constant $k$ is such solution that often works. So is setting it to $kt$ (where t is our independent variable), or $e^t$.

However, when the equation is homogeneous and contains the independent variable, some of the solutions seem to break down.

For example:

$$y''+ 6y = e^t$$ Or $$y''-4y'+8y = 2t^2$$

Now, the particular solution to the first equation can't be $y=k$. Since if $y=k$, then $y=\frac{e^t}{6}$, which is not a constant. How to think about finding particular solutions to the equations above?

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    $\begingroup$ do you intend $6^y$ or $6y$. One of these is more interesting than the other... $\endgroup$ – James S. Cook Feb 14 '16 at 13:28
  • $\begingroup$ @JamesS.Cook Well spotted... Fixed. $\endgroup$ – Dole Feb 14 '16 at 13:41
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    $\begingroup$ @Dole: Are you familiar with Undetermined Coefficients? $\endgroup$ – Moo Feb 14 '16 at 14:24
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    $\begingroup$ @Dole: In the second equation, I think you meant $4 y'$. $\endgroup$ – Moo Feb 14 '16 at 14:33

For the trial solution to work, the right hand side need to be of the sum-of-exponentials-with-polynomial-coefficients type. If one term of the sum is $p(t)e^{\lambda t}$ with $p(t)$ a polynomial, then the trial solution for that term is $t^\mu q(t)e^{\lambda t}$ where $\mu$ is the multiplicity of $\lambda$ as a root of the characteristic equation of the homogeneous part of the ODE and $q$ has the same degree as $p$ and the coefficients of $q$ are the to be determined parameters.

Note that due to $\cos t=\frac12(e^{it}+e^{-it})$ etc., also the trigonometric (and hyperbolic) sine and cosine fall under the exponential type.

For your example, that means that $Ae^t$ and $At^2+Bt+C$ are the trial solutions.


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