Solve the following differential equation x^2 d y d x =x^2+x y-y^… - Vidyadip

Question

Solve the following differential equation
$x^2 \frac{d y}{d x}=x^2+x y-y^2$

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Answer

$ x^2 \frac{ d y}{ d x}= x ^2+ xy - y ^2$
$\therefore \frac{ d y}{ d x}=1+\frac{y}{x}-\left(\frac{y}{x}\right)^2\ldots(i) $
Put $\frac{y}{x}= t$$\ldots(ii)$
$\therefore y = tx$
Differentiating w.r.t. $x$, we get
$\frac{ d y}{ d x}= t +x \frac{ dt }{ d x}\ldots(iii)$
Substituting (ii) and (iii) in (i), we get
$ t +x \frac{ dt }{ d x}=1+ t - t ^2$
$\therefore x \frac{ dt }{ d x}=1- t ^2$
$\therefore \frac{ dt }{1- t ^2}=\frac{ d x}{x} $
Integrating on both sides, we get
$ \int \frac{ dt }{1- t ^2}=\int \frac{ d x}{x}$
$\therefore \frac{1}{2} \log \left|\frac{1+t}{1-t}\right|=\log | x |+\log \left| c _1\right|$
$\therefore \log \left|\frac{1+\frac{y}{x}}{1-\frac{y}{x}}\right|=2 \log | x |+2 \log \left| c _1\right| $
$\therefore \log \left|\frac{x+y}{x-y}\right|=\log \left| x ^2\right|+\log \left| c _1{ }^2\right|$
$\therefore \log \left|\frac{x+y}{x-y}\right|=\log \left| c ^1 x ^2\right|$
$\therefore \frac{x+y}{x-y}= c _1{ }^2 x ^2$
$\therefore \frac{x+y}{x-y}= cx ^2, \text { where } c = c _1{ }^2$

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