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##### Quiz-3 / QUIZ3 5301 TWO-C

« Last post by**Jin Qin**on

*February 19, 2021, 06:30:49 PM*»

Hi, this is my answer for QUIZ3 TWO-C in section 5301. Hope this can help you out!

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Hi, this is my answer for QUIZ3 TWO-C in section 5301. Hope this can help you out!

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Everything is correct. You need to look carefully at limits in the integrals

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$\int_0^\infty$. I fixed it

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I am not sure what should this integral integrate over. Is it -∞ to +∞ or 0 to +∞?

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I believe as t->0^{+} and x>0, the integral I marked in red should be sqrt(pi), then U(x, t) should be 1/2. I do not know how we get 1.

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Indeed

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Should here be -2 instead of -25?

I have attached the picture below.

I have attached the picture below.

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Try to avoid high-riser notations. Several years ago I was a referee for a paper which used notations like this $\widehat{\dot{\widetilde{\mathcal{D}}}}$ and sometimes this little pesky dot was missing

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Indeed, this equation has $u=0$ as a solution but the notion of "homogeneou"s does not apply to nonlinear.

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$$\text{Find the general solutions to the following equation: }

u_{xxy}=x\cos(y)$$

\begin{gathered}

\nonumber

u_{xxy}=x\cos(y)\\

u_{xx}=x\sin(y) + f(x)\\

u_{x}=\frac{x^2}2\sin(y) + \tilde{f}(x) + h(y) ;\tilde{f}_x=f \\

u= \frac{x^3}6\sin(y) + \tilde{\tilde{f}}(x) + xh(y) + g(y);\tilde{\tilde{f}}_{xx}=f \\

\therefore u(x,y)=\frac{x^3}6\sin(y) + \tilde{\tilde{f}}(x) + xh(y) + g(y).\\

\text{Where $f(x),h(y),g(y)$ are arbitrary functions, and $\tilde{\tilde{f}}$ is twice differentiable.}

\end{gathered}

u_{xxy}=x\cos(y)$$

\begin{gathered}

\nonumber

u_{xxy}=x\cos(y)\\

u_{xx}=x\sin(y) + f(x)\\

u_{x}=\frac{x^2}2\sin(y) + \tilde{f}(x) + h(y) ;\tilde{f}_x=f \\

u= \frac{x^3}6\sin(y) + \tilde{\tilde{f}}(x) + xh(y) + g(y);\tilde{\tilde{f}}_{xx}=f \\

\therefore u(x,y)=\frac{x^3}6\sin(y) + \tilde{\tilde{f}}(x) + xh(y) + g(y).\\

\text{Where $f(x),h(y),g(y)$ are arbitrary functions, and $\tilde{\tilde{f}}$ is twice differentiable.}

\end{gathered}