I already solved for the homogeneous one, but I'm still looking for the particular solution of the differential equation:


The homogeneous solutions of this system are $\mathrm{e}^{-x}$ and $\mathrm{e}^{-2x}$. I've tried the substitution

$$y_p(x)=v(x)\mathrm{e}^{-2x},$$ which yielded the differential equation:


After that I reduced the order by $p=v'$ which gives:


Now this equation still seems hard to solve. I was wondering if there were easier/better substitutions to make ?

  • 1
    $\begingroup$ I have the feeling that it should be $v''-v'=\exp(e^x+2x)$ $\endgroup$ – Claude Leibovici Oct 10 '14 at 9:58
  • $\begingroup$ Have you learned the Wronskian? Also, that last DE can be solved easily with an integrating factor. $\endgroup$ – UserX Oct 10 '14 at 10:00
  • $\begingroup$ @UserX, the Wronskian is known to me :). $\endgroup$ – Nick Oct 10 '14 at 11:01

We first observe that $$ y=\frac{1}{(n+1)(n+2)}\mathrm{e}^{nx}, $$ is a particular solution of $$ y''+3y'+2y=\mathrm{e}^{nx}. $$ Hence $$ \sum_{n=0}^\infty \frac{1}{(n+1)(n+2)n!}\mathrm{e}^{nx}=\mathrm{e}^{-2x}\left(\exp(\mathrm{e}^x)-1-\mathrm{e}^x\right) $$ is a particular solution of $$ y''+3y'+2y=\sum_{n=0}^\infty\frac{1}{n!}\mathrm{e}^{nx}=\exp(\mathrm{e}^x). $$

Note. However, if one wants to be rigorous, a verification would be the right thing to do.

  • $\begingroup$ this includes the complentary solution, because using my method above I have found a partial amount of your answer. However, when added with my complementary solution, I get the same answer as you do. My $c_1$ and $c_2$ are arbitrary constants. Meaning, you have found the general solution, not the particular solution. $\endgroup$ – Varun Iyer Oct 10 '14 at 10:38
  • $\begingroup$ @YiorgosS.Smyrlis, that's a very nice trick :) $\endgroup$ – Nick Oct 10 '14 at 11:04

This solution might be more time-comsuming, but it can work.

We can use a method called variation of parameters.

If we take our characteristic polynomial:

$$y_c(t) = r^2+3r+2 = (r+2)(r+1)$$

Now we have that our complementary solution is:

$$y_c(t) = c_1e^{-2t} + c_2e^{-t}$$

Now our solution to this equation is $e^{-2t}$ and $e^{-t}$

We can use this to find our Wronskian

$$W = \begin{vmatrix}e^{-2t} &e^{-t} \\ -2e^{-2t} & -e^{-t} \end{vmatrix} = -e^{-3t}+2e^{-3t} = e^{-3t}$$

Now we can find our particular solution.

$$y_p(t) = -e^{-2t}\int\frac{e^{-t}e^{e^t}}{e^{-3t}}dt + e^{-t}\int\frac{e^{-2t}e^{e^t}}{e^{-3t}}dt$$

Solving this will get you the particular solution.


Solving for the particular we get

$$y_p(t) = -e^{-2t}\left(e^{e^t}(e^{t}-1)\right) + e^{-t}e^{e^t}$$ $$y_p(t) = -e^{-2t}(e^{e^t}e^{t}-e^{e^t}) + e^{-t}e^{e^t} = e^{-2t}e^{e^t}$$

If you need the general solution, add both the complementary and the particular solution together.


I believe Claude's comment is correct an that differential equation is not pretty. So I'll attempt a different approach, attempting to solve for the general solution.

Since the characteristic polynomial is $(s+2)(s+1)$, it appears integrating factors of $e^x$ or $e^{2x}$ will work. $e^x$ will result in easier integration. So we let

$$z=y'+2y$$ $$y''+3y'+2y=(y'+2y)'+y'+2y=z'+z=e^{e^x}$$ $$e^xz'+e^xz=(e^xz)'=e^xe^{e^x}$$ $$e^xz=e^{e^x}+k_1$$ $$e^x(y'+2y)=e^{e^x}+k_1$$ $$e^{2x}(y'+2y)=(e^{2x}y)'=e^xe^{e^x}+k_1e^x$$ $$e^{2x}y=e^{e^x}+k_1e^x+k_2$$ $$y=e^{e^x-2x}+k_1e^{-x}+k_2e^{-2x}$$


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.