Toronto Math Forum

APM346-2016F => APM346--Tests => Q1 => Topic started by: Victor Ivrii on September 29, 2016, 09:30:39 PM

Title: Q1-P3
Post 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{eq-1}\\
&u|_{x=0}=y.\label{eq-2}
\end{aligned} \right.
\end{equation}

Title: Re: Q1-P3
Post by: Roro Sihui Yap on September 29, 2016, 09:32:27 PM

\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, $u = (y - 3x)e^x$