2
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @Dole: Are you familiar with Undetermined Coefficients? $\endgroup$ – Moo Feb 14 '16 at 14:24
  • 1
    $\begingroup$ @Dole: In the second equation, I think you meant $4 y'$. $\endgroup$ – Moo Feb 14 '16 at 14:33
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.