<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}

Solve the following non-linear differential equations

\begin{description}
\item[(a)] $\ds x^2y^2y'-(1+x)(1+y)=0$
\item[(b)] $\ds y'=2e^{\frac{y}{x}}+\frac{y}{x}$
\item[(c)] $\ds (y\sec^2x+\sec x \tan x)+(\tan x+2y)y'=0$
\item[(d)] $\ds (x^2-1)y'+2xy^2=0, \qquad y(\sqrt 2)=\frac{1}{2}$
\end{description}


\vspace{.5in}

{\bf Answer}

\begin{description}
\item[(a)] A separable ODE
\begin{eqnarray*} x^2y^2y' -(1+x)(1+y) &amp; = &amp; 0 \\ \Rightarrow
x^2y^2 \frac{dy}{dx} &amp; = &amp; (1+x)(1+y) \\ \Rightarrow \int
\frac{y^2}{1+y}dy &amp; = &amp; \int \frac{1+ x}{x^2}dx \\ {\rm Now\ \ \ }
\int\frac{y^2}{1+y} dy &amp; = &amp; \int \left(y - 1 +
\frac{1}{y+1}\right)dy  \\ &amp; = &amp; \frac{1}{2}y^2 - y + \ln(y+1) \\
{\rm so\ \ \ } \frac{1}{2}y^2 - y + \ln (y +1) &amp; = &amp; \int \left(
\frac{1+x}{x^2} \right) dx \\ &amp; = &amp; -\frac{1}{x} + \ln x + c \\
{\rm i.e.\ \ \ }\frac{1}{2}y^2 - y + \ln (y +1) &amp; = &amp; -\frac{1}{x}
+ \ln x + c \end{eqnarray*}


\item[(b)] A homogeneous ODE in $x$ and $y$
$$\frac{dy}{dx} = 2e^{\frac{y}{x}} + \frac{y}{x}\hspace{.2in}
(1)$$ Let $\ds \frac{y}{x} = v$, $y = vx$, $\ds \frac{dy}{dx} = v
+ x\frac{dv}{dx}$

So (1) becomes \begin{eqnarray*}  v + x \frac{dv}{dx} &amp; = &amp; 2e^v +
v \\ \Rightarrow x \frac{dv}{dx} &amp;=&amp; 2e^v \\ \Rightarrow \int
e^{-v} &amp; = &amp; 2\int \frac{dx}{x} \\ -e^{-v} &amp; = &amp; 2 \ln x + c \\
-e^{-\frac{y}{x}} &amp; = &amp; 2 \ln x + c \end{eqnarray*}


\item[(c)] An exact ODE $$\ds (y\sec^2x+\sec x \tan x)+(\tan
x+2y)\frac{dy}{dx}=0$$

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

$\ds \frac{\pl P}{\pl y} = \sec^2x = \frac{\pl Q}{\pl x}$ so the
equation is exact.

Now we need to find $F(x,y)$ such that \begin{eqnarray} \frac{\pl
F}{\pl x} = P &amp; = &amp; y\sec^2x + \sec x \tan x \\ \frac{\pl F}{\pl
y} = Q &amp; = &amp; \tan x + 2y \end{eqnarray}

Integrating (2) gives $\ds F = y\tan x + y^2 + f(x)$
\hspace{1.2in} (3)

Differentiating (3) w.r.t. $x$ gives   $\ds \frac{\pl F}{\pl x} =
y \sec^2x + f'(x)$\hspace{.7in}(4)

So
\begin{eqnarray*}f'(x) &amp; = &amp; \sec x \tan x \\\Rightarrow f(x) &amp; = &amp; \sec x +c
\\ \Rightarrow F &amp; = &amp; y \tan x + y^2 + \sec x + c \end{eqnarray*}

and so the solution is $F(x,y) =$constant $${\rm i.e.\ \ \ }  y
\tan x + y^2 + \sec x + c = K$$

\item[(d)] \begin{eqnarray*}  (x^2-1)y'+2xy^2 &amp; = &amp; 0 \\
\Rightarrow (x^2 - 1) \frac{dy}{dx} &amp; = &amp; -2xy^2 \\ \Rightarrow  -
\int\frac{dy}{y^2} &amp; = &amp; \int \frac{2x }{x^2 - 1}dx \\ \Rightarrow
\frac{1}{y} &amp; = &amp; \ln(x^2 - 1) + C \end{eqnarray*} when $x = \sqrt
2, y = \frac{1}{2} \Rightarrow 2 = C$ so the answer is $$ y =
\ln(x^2 - 1) + 2$$

\end{description}



\end{document}
</pre></body></html>