Author Topic: Q1-T5102-P1  (Read 3124 times)

Victor Ivrii

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Q1-T5102-P1
« on: January 25, 2018, 08:26:00 AM »
Find the general solutions to the following equation:
$$
u_{xy}=2u_x.
$$

Elliot Jarmain

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Re: Q1-T5102-P1
« Reply #1 on: January 25, 2018, 08:29:42 AM »
\begin{equation*}
   u_{xy} = 2u_x
\end{equation*}

Let $v = u_x$:

\begin{equation*}
   v_y = 2v
\end{equation*}

Solving this equation:

\begin{gather*}
   \frac{d v}{d y} = 2v\\
   \implies  \int{\frac{d v}{v}}
                = \int{2 \, d y} \\
   \implies ln|v| = 2y + \phi(x)\\
   \implies v = \pm e^{\phi(x)}e^{2y}\\
   \implies v = \varphi'(x) e^{2y}\\
\end{gather*}

Plugging $u_x$ back in for $v$:
\begin{gather*}
   \implies  u_x = \varphi'(x) e^{2y} \\
   \implies u = \varphi(x) e^{2y} + \psi(y)
\end{gather*}

Ruite Xu

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Re: Q1-T5102-P1
« Reply #2 on: January 25, 2018, 08:48:59 AM »
Since we have $u_{xy}=2u_x$
Let $v = u_x$
Now we have : $v_y = 2v $
which implies $v= \alpha'(x)e^{2y}$
integral with respect to x: $\int v dx = \int \alpha'(x)e^{2y} dx$
                               $ u = \alpha(x)e^{2y} + \phi(y)$

Check: $u_x = \alpha'(x)e^{2y} + 0$
$u_{xy} = 2\alpha'(x)e^{2y} = 2u_x$
« Last Edit: January 25, 2018, 08:53:33 AM by Ruite Xu »

Victor Ivrii

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Re: Q1-T5102-P1
« Reply #3 on: January 25, 2018, 09:00:58 AM »
Try better formatting. Also \ln