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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 :)

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  • $\begingroup$ 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
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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

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

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  • $\begingroup$ 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

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