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I'm having problems finding the final solution $u(x,t)$ to this differential equation.

$$u_n(x,t) = \sin(nπx)[C\cos(nπt) + D\sin(nπt)]$$

Conditions: $$u(x,0) = \sin(πx)$$ $$u'_t(x,0) = \sin(2πx)$$

Can somebody help me?

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  • $\begingroup$ Unless there's a typo, you should have no problem combining $\cos(nπt)$ and $\cos(nπt)$ $\endgroup$ – Kaynex Jul 11 '18 at 17:16
  • $\begingroup$ That's right. It should be sin $\endgroup$ – emee Jul 11 '18 at 17:24
  • $\begingroup$ Where is your differential equation? $\endgroup$ – Nosrati Jul 11 '18 at 17:36
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By looking at the initial conditions, it's clear that the coefficients are only non-zero for $n=1$ and $n=2$, so we can simplify

$$ u(x,t) = \sin(\pi x)\big[C_1\cos(\pi t) + D_1\sin(\pi t)\big] +\sin(2\pi x)\big[C_2\cos(2\pi t) + D_2\sin(2\pi t)\big] $$

Matching the conditions

\begin{align} u(x,0) &= C_1\sin(\pi x) + C_2\sin(2\pi x) = \sin(\pi x) &&\implies C_2 = 0, C_1 = 1 \\ u_t(x,0) &= \pi D_1\sin(2\pi x) + 2\pi D_2\sin(2\pi x) = \sin(2\pi x) &&\implies D_1 = 0, 2\pi D_2 = 1 \end{align}

So the final solution is

$$ u(x,t) = \sin(\pi x)\cos(\pi t) + \frac{1}{2\pi}\sin(2\pi x)\sin(2\pi t) $$

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