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I am given a nonhomogeneous differential equation:

$$y''+4y'+3y=g(x)$$

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

After working through the problem, I have

$$y_c(x)=C_1e^{-3x}+C_2e^{-x}$$

(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?

Thanks!

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  • $\begingroup$ 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
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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 $(y_{1}+y_{2})^{(n)}(x) = y_{1}^{(n)}(x) + y_{2}^{(n)}(x)$.

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

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