- #1
AntSC
- 65
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Can someone check my working. I don't understand why i am getting different answers?
[tex]u(x,t)=\frac{{e}^{-\frac{x^2}{4Dt}}}{\sqrt{4Dt}}[/tex]
Differentiate w.r.t 't' by quotient rule:
[tex]\frac{\partial u}{\partial t}=\left[ \frac{1}{\sqrt{4Dt}}\cdot \frac{x^2}{4Dt^2}\cdot {e}^{-\frac{x^2}{4Dt}}-{e}^{-\frac{x^2}{4Dt}}\cdot \frac{1}{\sqrt{4D}}\cdot \frac{-1}{2t\sqrt{t}} \right]\frac{1}{4Dt}[/tex]
[tex]\frac{\partial u}{\partial t}=\left[ \frac{x^2}{4Dt^2}u(x,t)+\frac{1}{2t}u(x,t)\right]\frac{1}{4Dt}[/tex]
For the product rule, write as -
[tex]u(x,t)={\left(4D \right)}^{-\frac{1}{2}}{t}^{-\frac{1}{2}}{e}^{-\frac{x^2}{4Dt}}[/tex]
Differentiate w.r.t 't' by product rule:
[tex]\frac{\partial u}{\partial t}=\frac{1}{\sqrt{4D}}\left[ {e}^{-\frac{x^2}{4Dt}}\cdot \frac{-1}{2t\sqrt{t}}+\frac{1}{\sqrt{t}}\cdot \frac{x^2}{4Dt^2}\cdot {e}^{-\frac{x^2}{4Dt}} \right][/tex]
[tex]\frac{\partial u}{\partial t}=\frac{x^2}{4Dt^2}u(x,t)-\frac{1}{2t}u(x,t)[/tex]
This is clearly not the same.
What am i doing wrong?
[tex]u(x,t)=\frac{{e}^{-\frac{x^2}{4Dt}}}{\sqrt{4Dt}}[/tex]
Differentiate w.r.t 't' by quotient rule:
[tex]\frac{\partial u}{\partial t}=\left[ \frac{1}{\sqrt{4Dt}}\cdot \frac{x^2}{4Dt^2}\cdot {e}^{-\frac{x^2}{4Dt}}-{e}^{-\frac{x^2}{4Dt}}\cdot \frac{1}{\sqrt{4D}}\cdot \frac{-1}{2t\sqrt{t}} \right]\frac{1}{4Dt}[/tex]
[tex]\frac{\partial u}{\partial t}=\left[ \frac{x^2}{4Dt^2}u(x,t)+\frac{1}{2t}u(x,t)\right]\frac{1}{4Dt}[/tex]
For the product rule, write as -
[tex]u(x,t)={\left(4D \right)}^{-\frac{1}{2}}{t}^{-\frac{1}{2}}{e}^{-\frac{x^2}{4Dt}}[/tex]
Differentiate w.r.t 't' by product rule:
[tex]\frac{\partial u}{\partial t}=\frac{1}{\sqrt{4D}}\left[ {e}^{-\frac{x^2}{4Dt}}\cdot \frac{-1}{2t\sqrt{t}}+\frac{1}{\sqrt{t}}\cdot \frac{x^2}{4Dt^2}\cdot {e}^{-\frac{x^2}{4Dt}} \right][/tex]
[tex]\frac{\partial u}{\partial t}=\frac{x^2}{4Dt^2}u(x,t)-\frac{1}{2t}u(x,t)[/tex]
This is clearly not the same.
What am i doing wrong?
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