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Question Number 143920 by mohammad17 last updated on 19/Jun/21

Answered by Dwaipayan Shikari last updated on 19/Jun/21

$$a^{2}u={Ce}^{\pm ax+t}$$

Commented by mohammad17 last updated on 19/Jun/21

$$canyougivestebssir$$

Commented by Dwaipayan Shikari last updated on 19/Jun/21

$$I{don}^{\prime }tknowanysystematicwaytosolve$$$$P.D.E.Butiobservedthat$$$$Onecancheck\frac{\partial u}{\partial t}={Ce}^{\pm ax+t}and\frac{{\partial }^{2}u}{\partial x^2}={(\pm a)}^2{Ce}^{\pm ax+t}$$$$\Rightarrow \frac{{\partial }^{2}u}{\partial x^2}=a^2\frac{\partial u}{\partial t}$$

Answered by Olaf_Thorendsen last updated on 19/Jun/21

$$\frac{\partial u}{\partial t}=a^2\frac{{\partial }^{2}u}{\partial x^2}\left(1\right)$$$$\mathrm{Usually},\mathrm{we}\mathrm{try}\mathrm{to}\mathrm{find}\mathrm{such}\mathrm{a}\mathrm{solution}$$$$u(x,t)=\mathrm{A}\left(x\right)\mathrm{B}\left(t\right)$$$$\left(1\right):\mathrm{A}\left(x\right){\mathrm{B}}^{\prime }\left(t\right)=a^2{\mathrm{A}}^{″}\left(x\right)\mathrm{B}\left(t\right)$$$$a^2\frac{{\mathrm{A}}^{″}\left(x\right)}{\mathrm{A}\left(x\right)}=\frac{{\mathrm{B}}^{\prime }\left(t\right)}{\mathrm{B}\left(t\right)}\left(2\right)$$$$\mathrm{Necessarily}\left(2\right)=\mathrm{constant}$$$$\left(2\right):\frac{{\mathrm{B}}^{\prime }\left(t\right)}{\mathrm{B}\left(t\right)}=\mathrm{cst}=-{\alpha }^2$$$$\mathrm{B}\left(t\right)={\mathrm{B}}_0{e}^{-{\alpha }^{2}t}\left(3\right)$$$$\left(2\right):a^2\frac{{\mathrm{A}}^{″}\left(x\right)}{\mathrm{A}\left(x\right)}=\mathrm{cst}=-{\alpha }^2$$$$\mathrm{A}\left(x\right)=\lambda \cos \left(\frac{\alpha x}{a}\right)+\mu \sin \left(\frac{\alpha x}{a}\right)\left(4\right)$$$$$$$$\mathrm{Limit}\mathrm{conditions}:$$$$$$$$u(0,t)=0=\lambda {\mathrm{B}}_0{e}^{-{\alpha }^{2}t}$$$$\Rightarrow \lambda =0$$$$u(x,t)=\mu \sin \left(\frac{\alpha x}{a}\right){\mathrm{B}}_0{e}^{-{\alpha }^{2}t}={\mathrm{U}}_0\sin \left(\frac{\alpha x}{a}\right)e^{-{\alpha }^{2}t}$$$$$$$$u(l,t)=0={\mathrm{U}}_0\sin \left(\frac{\alpha l}{a}\right)e^{-{\alpha }^{2}t}$$$$\Rightarrow \frac{\alpha l}{a}=\pi$$$$u(x,t)={\mathrm{U}}_0\sin \left(\frac{\pi x}{l}\right)e^{-\frac{{\pi }^2{a}^2}{l^2}t}$$$$$$$$u(x,0)=3\sin \left(\frac{\pi x}{l}\right)={\mathrm{U}}_0\sin \left(\frac{\pi x}{l}\right)e^0$$$$\Rightarrow {\mathrm{U}}_0=3$$$$$$$$\mathrm{Finally}u(x,t)=3\sin \left(\frac{\pi x}{l}\right)e^{-\frac{{\pi }^2{a}^{2}t}{l^2}}$$

Commented by mohammad17 last updated on 20/Jun/21

$$thankyousir$$

Commented by Dwaipayan Shikari last updated on 20/Jun/21

$$Thankssir$$

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