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$u_t + 3u_x = 2t$, $u(x,0)=\sin(x/2)$.

I used the method of characteristics to get the answer, $u(x,t)=t^2 + 2\sin^{-1}(x-3t)$.

Does this satisfy the initial condition? I checked for the first equation and it does; however I do not think it satisfies the intial value when $t=0.$ Am I correct in saying so?

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  • $\begingroup$ Yes you are corect to saying that it does not satisfy the IC. It is clear when you put $t=0$ in the solution you have found. $\endgroup$
    – daulomb
    Jan 16 '14 at 3:48
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The solution is $u=\frac{2}{3}\cos(\frac{x}{2})+\sin(\frac{x-3t}{2})-\frac{2}{3}\cos(\frac{x-3t}{2})$.

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I think your u(x,t) is not correct: apart from some factors of 2 different, I have $ \sin $ and not $ \sin^{-1} $.

Also, worth recognizing that you can solve this with just change of variables, $ x \rightarrow x + 3 t $.

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answer for above question is $$u(x,t)=t^2+\sin(x−3t/2)$$

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  • $\begingroup$ This answer could be improved with an explanation. $\endgroup$
    – Null
    Oct 28 '14 at 16:07

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