Author Topic: TUT 0602 Quiz 2  (Read 467 times)

Yichen Ji

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TUT 0602 Quiz 2
« on: November 04, 2019, 06:40:19 PM »
Question: show that the given equation is not exact but becomes exact when multiplied by the given integrating factor.Then solve the equation.
The equation is:
\begin{equation*}
    x^2y^3+x(1+y^2)y'=0
\end{equation*}
The integrating factor $\mu(x,y)=\frac{1}{xy^3}$

Solution:
First,verify that the initial equation is not exact:
\begin{align*}
    M(x,y) &=x^2y^3 & N(x,y) &=x+xy^2\\
    M_y &=3x^2y^2 & N_x &=1+y^2\\
\end{align*}
$M_y \neq N_x$,the equation is not exact.
Next, multiplying $\mu$ on both sides,
\begin{align*}
    \mu x^2y^3+\mu x(1+y^2)y'=0\\
    x+\frac{1+y^2}{y^3}y'=0
\end{align*}
Denote $P(x,y)=x$ and $Q(x,y)=\frac{1+y^2}{y^3}$
Check exactness:
\begin{equation*}
    P_y=Q_x=0
\end{equation*}
The new equation is exact.
Now solve the equation:
\begin{align*}
    \frac{\partial\psi}{\partial x}=x\\
    \psi=\frac{1}{2}x^2+g(y)\\
    \frac{\partial\psi}{\partial y}=g'(y)=y^{-3}+y^{-1}\\
    g(y)=-\frac{1}{2}y^{-2}+ln|y|
\end{align*}
So the solution to the equation is
\begin{equation*}
    \psi(x,y)=\frac{x^2}{2}-\frac{1}{2y^2}+ln|y|=c
\end{equation*}
for $c$ being constant.