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Question-107941

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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
(∂θ/∂t)=nt^(n−1) e^((−r^2 )/(2t)) +t^n (r^2 /(2t^2 ))e^((−r^2 )/(2t)) =(nt^(n−1) +(r^2 /2)t^(n−2) )e^((−r^2 )/(2t)) (∗) and (∂θ/∂r)=t^n ((−2r)/(2t))e^((−r^2 )/(2t)) =−rt^(n−1) e^((−r^2 )/(2t)) and (1/r^2 )(∂/∂r)(r^2 (∂θ/∂r))=(1/r^2 ) (∂/∂r)(−r^3 t^(n−1) e^((−r^2 )/(2t)) )= (1/r^2 )(−3r^2 t^(n−1) e^((−r^2 )/(2t)) +r^4 t^(n−2) e^((−r^2 )/(2t)) )= (−3t^(n−1) +r^2 t^(n−2) )e^((−r^2 )/(2t)) (∗∗) and from (∗)=(∗∗) we have −3t^(n−1) +r^2 t^(n−2) =nt^(n−1) +(r^2 /2)t^(n−2) ⇒ (n+3)t^(n−1) +((r^2 /2)−r^2 )t^(n−2) =0⇒ (n+3)t^(n−1) −(r^2 /2)t^(n−2) =0⇒ (n+3)t−(r^2 /2)=0⇒n+3=(r^2 /(2t))⇒n=−3+(r^2 /(2t))
$$\frac{\partial\theta}{\partial{t}}={nt}^{{n}−\mathrm{1}} {e}^{\frac{−{r}^{\mathrm{2}} }{\mathrm{2}{t}}} +{t}^{{n}} \frac{{r}^{\mathrm{2}} }{\mathrm{2}{t}^{\mathrm{2}} }{e}^{\frac{−{r}^{\mathrm{2}} }{\mathrm{2}{t}}} =\left({nt}^{{n}−\mathrm{1}} +\frac{{r}^{\mathrm{2}} }{\mathrm{2}}{t}^{{n}−\mathrm{2}} \right){e}^{\frac{−{r}^{\mathrm{2}} }{\mathrm{2}{t}}} \:\left(\ast\right) \\ $$$${and} \\ $$$$\frac{\partial\theta}{\partial{r}}={t}^{{n}} \:\frac{−\mathrm{2}{r}}{\mathrm{2}{t}}{e}^{\frac{−{r}^{\mathrm{2}} }{\mathrm{2}{t}}} =−{rt}^{{n}−\mathrm{1}} {e}^{\frac{−{r}^{\mathrm{2}} }{\mathrm{2}{t}}} \\ $$$${and} \\ $$$$\frac{\mathrm{1}}{{r}^{\mathrm{2}} }\frac{\partial}{\partial{r}}\left({r}^{\mathrm{2}} \frac{\partial\theta}{\partial{r}}\right)=\frac{\mathrm{1}}{{r}^{\mathrm{2}} }\:\frac{\partial}{\partial{r}}\left(−{r}^{\mathrm{3}} {t}^{{n}−\mathrm{1}} {e}^{\frac{−{r}^{\mathrm{2}} }{\mathrm{2}{t}}} \right)= \\ $$$$\frac{\mathrm{1}}{{r}^{\mathrm{2}} }\left(−\mathrm{3}{r}^{\mathrm{2}} {t}^{{n}−\mathrm{1}} {e}^{\frac{−{r}^{\mathrm{2}} }{\mathrm{2}{t}}} +{r}^{\mathrm{4}} {t}^{{n}−\mathrm{2}} {e}^{\frac{−{r}^{\mathrm{2}} }{\mathrm{2}{t}}} \right)= \\ $$$$\left(−\mathrm{3}{t}^{{n}−\mathrm{1}} +{r}^{\mathrm{2}} {t}^{{n}−\mathrm{2}} \right){e}^{\frac{−{r}^{\mathrm{2}} }{\mathrm{2}{t}}} \:\:\:\left(\ast\ast\right) \\ $$$${and}\:{from}\:\left(\ast\right)=\left(\ast\ast\right)\:\:{we}\:{have} \\ $$$$−\mathrm{3}{t}^{{n}−\mathrm{1}} +{r}^{\mathrm{2}} {t}^{{n}−\mathrm{2}} ={nt}^{{n}−\mathrm{1}} +\frac{{r}^{\mathrm{2}} }{\mathrm{2}}{t}^{{n}−\mathrm{2}} \Rightarrow \\ $$$$\left({n}+\mathrm{3}\right){t}^{{n}−\mathrm{1}} +\left(\frac{{r}^{\mathrm{2}} }{\mathrm{2}}−{r}^{\mathrm{2}} \right){t}^{{n}−\mathrm{2}} =\mathrm{0}\Rightarrow \\ $$$$\left({n}+\mathrm{3}\right){t}^{{n}−\mathrm{1}} −\frac{{r}^{\mathrm{2}} }{\mathrm{2}}{t}^{{n}−\mathrm{2}} =\mathrm{0}\Rightarrow \\ $$$$\left({n}+\mathrm{3}\right){t}−\frac{{r}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{0}\Rightarrow{n}+\mathrm{3}=\frac{{r}^{\mathrm{2}} }{\mathrm{2}{t}}\Rightarrow{n}=−\mathrm{3}+\frac{{r}^{\mathrm{2}} }{\mathrm{2}{t}} \\ $$

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