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QUESTION

Use Laplace transforms to solve the partial differential equation
$$y_x+2xy_t=2x$$ given that $y(x,0)=y(0,t)=1$.

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ANSWER

\begin{eqnarray}
 y_x+2xy_t=2x\\
 y(x,0)=1\\
 y(0,t)=1
\end{eqnarray}
The Laplace transform of (1): $Y_x+2x(Y-sy(x,0)=\frac{2x}{s}\\
Y_x+2sxY=\frac{2x}{x}+2x\\
\frac{d}{dx}\left(Ye^{sx^2}\right)=e^{sx^2}\left(\frac{2x}{s}+2x\right)\\
Ye^{sx^2}=\int e
^{sx^2}\left(\frac{2x}{s}+2x\right)\,dx=e^{sx^2}\left(s^{-1}+s^{-2}\right)+C(s)\\
Y(x,s)=\frac{1}{s}+\frac{1}{s^2}+C(s)e^{-sx^2}$\\ To determine
C(s), take the Laplace transform of (3)
\begin{eqnarray*}
&amp;\Rightarrow&amp; Y(0,s)=\frac{1}{s}\\
&amp;\Rightarrow&amp;C(s)=-\frac{1}{s^2}\\
&amp;\Rightarrow&amp;Y(x,s)=\frac{1}{s}+\frac{1}{s^2}\left(1-e^{-sx^2}\right)\\
&amp;\Rightarrow&amp;y(x,t)=1+t-H(t-x^2)(t-x^2)\\
&amp;\Rightarrow&amp;y(x,t)=\left\{\begin{array}{l} 1+t\textrm{ for
}t&lt;x^2\\ 1+x^2\textrm{ for } t&gt;x^2\end{array}\right.
\end{eqnarray*}

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