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I have heat equation $$ u_t=u_xx \\ u(x,0)=xe^{\frac{-x^2}{3}} $$ I solve by the Poisson integral $$ u(x,t)=\frac{1}{2\sqrt{\pi t}} \int xe^{\frac{-x^2}{3}}e^{\frac{-(y-x)^2}{4t}} dx $$

I worked with him (see image), but I have no idea how to finish it. Can u help me with it? Thanks enter image description here

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$$A=\frac{(y-x)^2}{4 t}+\frac{x^2}{3}=\left(\frac{1}{4 t}+\frac{1}{3}\right) x^2-\frac{ y}{2 t}x+\frac{y^2}{4 t}$$ Complete the square to get $$\left(\frac{1}{4 t}+\frac{1}{3}\right)\Bigg[\left(x-\frac{3 y}{4 t+3}\right)^2 +\frac{12 t y^2}{(4 t+3)^2}\Bigg]$$ Lat $$a=\left(\frac{1}{4 t}+\frac{1}{3}\right)\qquad b=\frac{3 y}{4 t+3}\qquad c=\left(\frac{1}{4 t}+\frac{1}{3}\right)\frac{12 t y^2}{(4 t+3)^2}$$ there are simplifications to be done and you face $$\int x e^{-a(x-b)^2-c}\,dx=e^{-c}\int x e^{-a(x-b)^2}\,dx$$ Let $x=b+y$ to face two usual inetgrals.

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  • $\begingroup$ Thank you! But i have a question: is x=b+t correct? Maybe another variable instead of t? $\endgroup$ – Sergey Prikhodko Jan 17 at 10:05
  • $\begingroup$ @SergeyPrikhodko You are right ! It is fixed now (I hope !) $\endgroup$ – Claude Leibovici Jan 17 at 10:07

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