Author Topic: LEC0201-Retest-ALT-Y-Question1  (Read 234 times)

RunboZhang

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LEC0201-Retest-ALT-Y-Question1
« on: October 23, 2020, 04:18:51 PM »
$\textbf {Problem:} \\\\ $
$\text{(a) Find integrating factor and then a general solution of ODE} \\$
$\begin{gather}
(1+xy+y^{2}) + (1+xy+x^{2})y' = 0
\end{gather}$
$\text{(b) Also, find a solution satisfying y(1) = 1}\\\\$

$\textbf{Solution: } \\\\$
$\text{(a):}\\\\$
$\text{We have:}$

$
\begin{gather}
\begin{aligned}

M_y = x+2y \\\\
N_x = y+2x

\end{aligned}
\end{gather}
$

$\text{Since} M_y \ne N_x \ \text{, thus equation is not exact.}$

$
\begin{gather}
\begin{aligned}

\frac{N_x - M_y}{M \cdot x - N \cdot y} &{}= \frac{y+2x-x-2y}{x+x^{2}y+xy^{2}-y-xy^{2}-x^{2}y} \\\\
&{} = \frac{x - y}{x - y} \\\\
&{} = 1

\end{aligned}
\end{gather}
$

$\text{Now, we have our integrating factor } \mu \ \text{computed as follow: }$

$
\begin{gather}
\begin{aligned}

\mu &{} = e^{\int {1} \, dxy} \\\\
&{} = e^{xy}

\end{aligned}
\end{gather}
$

$\text{Now multiplying the original equation with our integrating factor, we have: }$

$
\begin{gather}
\begin{aligned}

e^{xy} + e^{xy} \cdot xy + e^{xy} \cdot y^{2} + (e^{xy} + e^{xy} \cdot xy + e^{xy} \cdot x^{2})y' = 0

\end{aligned}
\end{gather}
$

$\text{Lastly, solve for general solution F(x,y): }$

$
\begin{gather}
\begin{aligned}

F(x,y) &{} = \int_{0}^{x} (e^{xy} + e^{xy} \cdot xy + e^{xy} \cdot y^{2}) \, dx + h'(y) \\\\
&{} = x \cdot e^{xy} + y \cdot e^{xy} = c

\end{aligned}
\end{gather}
$



$\text{(b):}\\\\$
$\text{By plugging in initial value condition, we have: }$
$
\begin{gather}
\begin{aligned}

y(1) = 1 \\\\
c = e + e = 2e\\\\

\implies x \cdot e^{xy} + y \cdot e^{xy} = 2e

\end{aligned}
\end{gather}
$