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is there a way to solve this equation using wolfram alpha:

$$\frac{\partial u}{\partial t}-\frac{\partial ^2u}{\partial x^2}=0 $$ and $u(t,0)=u(t,\pi)=0$, $u(0,x)=2\sin(2x)-3\sin(3x)$

Thanks all

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  • $\begingroup$ I’d seen it but still was having problem how to input my equation $\endgroup$ – user114138 Feb 23 at 16:58
  • $\begingroup$ Is that the correct equation? Because I see no dependence on $t$ at all. Viewed only as a function of $x$, $u' = u"$ which has exponential solutions that can't satisfy the boundary conditions. $\endgroup$ – RobertTheTutor Feb 23 at 17:09
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    $\begingroup$ I suspect it should be $$\frac{\partial u}{\partial t} - \frac{\partial^2 u}{\partial x^2} = 0$$ $\endgroup$ – Robert Israel Feb 23 at 17:19
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    $\begingroup$ $u=2e^{-4t}\sin(2x)-3e^{-9t}\sin(3x)$. $\endgroup$ – JJacquelin Feb 23 at 18:27
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    $\begingroup$ I was unable to get WolframAlpha to do it either (and I'm pretty good with both Mathematica and WolframAlpha). $\endgroup$ – Patrick Stevens Feb 23 at 18:59
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Try the Wolfram command

DSolve[
     {D[u[t,x], t]==D[u[t,x], {x,2}],
     u[0,x]==2Sin[2 x]-3Sin[3 x],
     u[t,0]==0, u[t,Pi]==0}, u, {{t,0,Infinity}, {x,0,Pi}}]

which returns

{{u->Function[{t,x},(2*E^(5*t)*Sin[2*x]- 3*Sin[3*x])/E^(9*t)]}}

You may still be able to get this somehow using Wolfram|Alpha but is it worth the effort?

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