Toronto Math Forum
APM3462016F => APM346Tests => Q1 => Topic started by: Victor Ivrii on September 29, 2016, 09:30:39 PM

Find the solution of
\begin{equation}
\left\{\begin{aligned}
&u_x+3u_y=u,\label{eq1}\\
&u_{x=0}=y.\label{eq2}
\end{aligned} \right.
\end{equation}

\begin{equation} \frac{dx}{1} = \frac{dy}{3} = \frac{du}{u} \end{equation}
From $\frac{dx}{1} = \frac{dy}{3} $, $3x  y = C_1$
From $\frac{dx}{1} = \frac{du}{u}$, $\ln u = x + ln C_2 $
So $u = C_2e^x = \phi(3x  y)e^x $
Since $u_{x=0} = y$,
$\phi(y) = y $ which means $ \phi(z) = z $
Therefore, $u = (y  3x)e^x$