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Question Number 107941 by mohammad17 last updated on 13/Aug/20

Commented by kaivan.ahmadi last updated on 13/Aug/20

$$\frac{\partial \theta }{\partial t}={nt}^{n-1}{e}^{\frac{-r^2}{2t}}+t^n\frac{r^2}{2t^2}{e}^{\frac{-r^2}{2t}}=({nt}^{n-1}+\frac{r^2}{2}{t}^{n-2})e^{\frac{-r^2}{2t}}(\ast )$$$$and$$$$\frac{\partial \theta }{\partial r}=t^n\frac{-2r}{2t}{e}^{\frac{-r^2}{2t}}=-{rt}^{n-1}{e}^{\frac{-r^2}{2t}}$$$$and$$$$\frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2\frac{\partial \theta }{\partial r}\right)=\frac{1}{r^2}\frac{\partial }{\partial r}(-r^3{t}^{n-1}{e}^{\frac{-r^2}{2t}})=$$$$\frac{1}{r^2}(-3r^2{t}^{n-1}{e}^{\frac{-r^2}{2t}}+r^4{t}^{n-2}{e}^{\frac{-r^2}{2t}})=$$$$(-3t^{n-1}+r^2{t}^{n-2})e^{\frac{-r^2}{2t}}(\ast \ast )$$$$andfrom(\ast )=(\ast \ast )wehave$$$$-3t^{n-1}+r^2{t}^{n-2}={nt}^{n-1}+\frac{r^2}{2}{t}^{n-2}\Rightarrow$$$$(n+3)t^{n-1}+(\frac{r^2}{2}-r^2)t^{n-2}=0\Rightarrow$$$$(n+3)t^{n-1}-\frac{r^2}{2}{t}^{n-2}=0\Rightarrow$$$$(n+3)t-\frac{r^2}{2}=0\Rightarrow n+3=\frac{r^2}{2t}\Rightarrow n=-3+\frac{r^2}{2t}$$

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