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If \( v=\frac{c}{\sqrt{t}} e^{-x^{2} / 4 a^{2} t} \) then show that \( \frac{\partial v}{\partial t}=a^{2} \frac{\partial^{2} v}{\partial x^{2}} \) |
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Answer» v = \(\frac{c}{\sqrt t}e^{\frac{-x^2}{4a^2t}}\) \(\frac{\partial v}{\partial x}=\frac{-2x}{4a^2t}\frac{c}{\sqrt t}e^{\frac{-x^2}{4a^2t}}\) \(=\frac{-2x}{4a^2t}v\) \(\frac{\partial^2v}{\partial x^2}=\frac{-2v}{4a^2t}-\frac{-2x}{4a^2t}.\frac{\partial v}{\partial x}\) \(=\frac{-2v}{4a^2t}+\frac{2x}{4a^2t}.\frac{2x}{4a^2t}v\) \(=\frac{-2v}{4a^2t}(1-\frac{2x^2}{4a^2t})\) \(a^2\frac{d^2v}{dx^2}=\frac{-v}{2t}(1-\frac{x^2}{2a^2t})\)-----(i) \(\frac{\partial v}{\partial t}=\frac{c}{\sqrt t}e^{-\frac{x^2}{4a^2t}}\) x \(\frac{x^2}{4a^2t^2}-\frac{c}{2t^{3/2}}e^{-\frac{x^2}{4a^2t}}\) \(=\frac{x^2v}{4a^2t^2}-\frac{v}{2t}\) \(= -\frac{v}{2t}(1-\frac{x^2}{2a^2t})\) \(=a^2\frac{\partial^2v}{\partial x^2}\) (From (i)) |
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