<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\newcommand{\ud}{\underline}
\newcommand{\lin}{\int_0^\infty}
\parindent=0pt
\begin{document}


{\bf Question}

Find an integrating factor and hence solve the equation
$$\left(2xy+x^2y+\frac{1}{3}y^3\right)dx+(x^2+y^2)dy=0$$



\vspace{.5in}

{\bf Answer}

$$\left(2xy+x^2y+\frac{1}{3}y^3\right)dx+(x^2+y^2)dy=0$$

$$P = 2xy+x^2y+\frac{1}{3}y^3  \hspace{.3in} Q = x^2+y^2$$

$$\frac{\pl P}{\pl y} = 2x + x^2 + y^2 \hspace{.3in} \frac{\pl
Q}{\pl x} = 2x$$

So $\ds \frac{\pl P}{\pl y} \not= \frac{\pl Q}{\pl x}$ so not
exact.

However it fails to be exact by $\ds \frac{\pl P}{\pl y} -
\frac{\pl Q}{\pl x} = x^2 + y^2 = Q$ so multiply by $e^x$

$$\left(2xy+x^2y+\frac{1}{3}y^3\right)e^xdx+(x^2+y^2)e^xdy=0$$

NOW

$$\tilde{P} = \left(2xy+x^2y+\frac{1}{3}y^3\right)e^x
\hspace{.3in} \tilde{Q} = (x^2+y^2)e^x$$

$$ \frac{\pl\tilde{P}}{\pl y} = (2x + x^2 + y^2)e^x \hspace{.3in}
\frac{\pl\tilde{Q}}{\pl x} = (2x + x^2 + y^2)e^x$$

So  $\ds \frac{\pl\tilde{P}}{\pl y} = \frac{\pl \tilde{Q}}{\pl x}$
and the equation is now exact.

So try and find $F(x,y)$ such that:

\begin{eqnarray*} \frac{\pl F}{\pl x} = \tilde{P} &amp; = &amp; (2xy +
x^2 y + \frac{1}{3}y^3)e^x \hspace{.5in}(1) \\ \frac{\pl F}{\pl y}
= \tilde{Q} &amp; = &amp; (x^2 + y^2) e^x\hspace{1.2in}(2) \end{eqnarray*}

Integrating (2) gives \begin{eqnarray*} F &amp; = &amp; (x^2y +
\frac{1}{3}y^3) e^x + f(x) \\ \frac{\pl F}{\pl x} &amp; = &amp; (2xy +
x^2y + \frac{1}{3}y^3)e^x + f'(x) \\ &amp; \Rightarrow &amp;  f'(x) = 0
\Rightarrow f(x) = c \\ {\rm so\ \ } F  &amp; = &amp; (x^2y +
\frac{1}{3}y^3)e^x + c \end{eqnarray*} and the solution is $F(x,y)
=$ constant.  $${\rm i.e.\ \ \ } \left(x^2y +
\frac{1}{3}y^3\right)e^x = K$$



\end{document}
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