I am given a nonhomogeneous differential equation:


where $g(x)=3 \sin 2x$.

After working through the problem, I have


(I was to find a general solution for which $g(x)=0$)

$$y_p(x)=-(24/65) \cos 2x-(3/65) \sin 2x. $$

(On this part, I was given $y_p(x)=A \cos 2x+ B \sin 2x$)

Now I'm stuck. How do I verify that $y_c(x)+y_p(x)$ is a solution to the differential equation?


  • $\begingroup$ Substitute it into the ODE. Take $y = y_h + y_p$, check $y'' + 4y' + 3 y$ and make sure it yields $3 \sin(2x)$ by substituting it back into the ODE and make sure both sides equate. Also, your solution is correct, but do as I said to make sure you understand. $\endgroup$ – Amzoti Dec 13 '13 at 22:12

Well, your solution $y_{c}(x)$ satisfies the problem $y'' + 4y' + 3y = 0$ and $y_{p}(x)$ satisfies the problem $y'' + 4y'+ 3y = g(x)$. So, $(y_{c}+y_{p})'' + 4(y_{c}+y_{p})' + 3(y_{c}+y_{p}) = [y_{c}'' + 4y_{c}' + 3y_{c}] + [y_{p}'' + 4y_{p}' + 3y_{p}] = 0 + g(x) = g(x)$. Hence, $y(x)$ satisfies the ODE. Note that derivatives of any order are additive; that is $(y_{1}+y_{2})^{(n)}(x) = y_{1}^{(n)}(x) + y_{2}^{(n)}(x)$.

| cite | improve this answer | |
  • 1
    $\begingroup$ That makes so much sense! Thanks for your help! $\endgroup$ – westhe32nd Dec 13 '13 at 22:22

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.