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Consider the following ODE:

$$\frac{d^2 x}{dt^2} + kx = f(t)$$

where $$f(t) = \frac{1}{2} +\sum_{n = 1}^{\infty}{ \frac{-4}{n \pi} \sin\left(\frac{n \pi t}{2}\right)}$$ was derived using fourier series.

I have to find a particular solution. The sum symbol confuses me on this one but I will treat it as one does normally when looking to find a particular solution.

My thoughts:

Guess the solution: $$ x(t) = A_1 t+ A_0 + \sum_{n = 1}^{\infty}{ B \sin\left(\frac{n \pi t}{2}\right) + C \cos\left(\frac{n \pi t}{2}\right)} $$

After taking the second derivative and matching coefficients I arrive to the conclusion that the coefficients of the particular solution are:

$$ A_1 = 0 $$

$$ A_0 = \frac{1}{2k} $$

$$ C = 0 $$

$$ B = - \frac{4}{(k - m (\frac{n \pi}{2})^2)n \pi} $$

Assuming the first and second derivatives are correct is this a correct approach?

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  • $\begingroup$ Looks ok to me. I guess you have $k>0$ You can also set $k=\lambda ^2$ and use variation of parameters. $\endgroup$ – Matematleta Dec 12 '15 at 16:17
  • $\begingroup$ @Chilango Undetermined comes easier to me. Relieved to know that its right :) Thanks! $\endgroup$ – DoubleOseven Dec 12 '15 at 16:20
  • $\begingroup$ if you use variation of parameters you will see the integrals are all easy because they are of the form sin/sin, sin/cos, etc $\endgroup$ – Matematleta Dec 12 '15 at 16:27
  • $\begingroup$ @Chilango I was thinking about $B$. Should I use the sum operator on when determining the coefficients? $\endgroup$ – DoubleOseven Dec 13 '15 at 12:14
  • $\begingroup$ is your diff eq an initial value problem?A boundary value problem? $\endgroup$ – Matematleta Dec 13 '15 at 15:41

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