For the differential equation $y'' + 5y' + 6y = 2\cos(x)$, a complementary solution is $y_c = c_1e^{-3x} + c_2e^{-2x}$ and a particular solution is $y_p = \frac{1}{5}(\cos(x) + \sin(x))$. Write a general solution to the problem, and then find a solution satisfying the initial conditions $y(0) = 0$, $y'(0) = 1$.

I'm having trouble starting. I thought that maybe taking the derivative of the particular solution would help:

$y_p' = \frac{1}{5}(\cos(x)-\sin(x))$

$y_p'' = \frac{1}{5}(-\cos(x)-\sin(x))$

But I'm not sure how I can use this to arrive at a general solution. Any help?

Thanks in advance :)

  • $\begingroup$ Type "\cos" and "\sin" $\endgroup$ – Stefan Smith Oct 3 '12 at 22:23
  • $\begingroup$ Thanks. (more characters) $\endgroup$ – Stefan Smith Oct 3 '12 at 23:49

Write your general solution as $ y_g(x) = y_c(x) + y_p(x) $, then use the initial conditions to find $c_1,c_2$. Note that, you have two constants to determine, so you need two equations in $c_1,c_2$. That means, you will use $y_g(x)$ and $y'_g(x)$. Once you find $c_1$ and $c_2$ plug them back in $y_g(x)$. The final answer is

$$ y(x) = -\frac{2}{5}\,{{\rm e}^{-3\,x}}+\frac{1}{5}\,{{\rm e}^{-2\,x}}+\frac{1}{5}\,\cos \left(x \right) + \frac{1}{5} \,\sin \left( x \right)$$

  • $\begingroup$ And of course you can always check your answer by seeing that it satisfies the initial conditions. $\endgroup$ – Gerry Myerson Oct 4 '12 at 7:46

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.