Solve the following differential equation y^2dx + (xy + x^2) dy =… - Vidyadip

Question

Solve the following differential equation $y^2dx + (xy + x^2) dy = 0$

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Answer

$ y ^2 d x+\left( xy + x ^2\right) d y=0$
$\therefore y^2+\left(x y+x^2\right) \frac{ d y}{ d x}=0$
$\therefore\left(x y+x^2\right) \frac{ d y}{ d x}=- y ^2$
$\therefore \frac{ d y}{ d x}=\frac{-y^2}{x y+x^2}\ldots(i) $
Put $y=t x$$\ldots(ii)$
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}=\frac{- t ^2 x^2}{x( t x)+x^2}$
$\therefore t +x \frac{ dt }{ d x}=\frac{- t ^2 x^2}{x^2( t +1)}$
$\therefore t +x \frac{ dt }{ d x}=\frac{- t ^2}{1+ t }$
$\therefore x \frac{ dt }{ d x}=\frac{- t ^2}{1+ t }- t$
$=\frac{- t ^2- t - t ^2}{1+ t } $
$ =\frac{-2 t ^2- t }{1+ t }$
$\therefore \frac{1+ t }{ t ^2+ t } dt =-\frac{ d x}{x} $
Integrating on both sides, we get
$ \int \frac{1+ t }{2 t ^2+ t } dt =-\int \frac{ d x}{x}$
$\therefore \int \frac{(2 t +1)- t }{ t (21 t +1)} dt =-\int \frac{ d x}{x}$
$\therefore \int\left(\frac{1}{ t }-\frac{1}{2 t +1}\right) dt =-\int \frac{ d x}{x}$
$\therefore \int \frac{1}{ t } dt -\frac{1}{2} \int \frac{2}{2 t +1} dt =-\int \frac{ d x}{x}$
$\therefore \log | t |-\frac{1}{2} \log |2 t +1|=-\log | x |+\log | c |$
$\therefore \log \left|\frac{y}{x}\right|-\frac{1}{2} \log \left|2\left(\frac{y}{x}\right)+1\right|=-\log | x |+\log | c |$
$\therefore \log |y|-\log |x|-\frac{1}{2} \log \left|\frac{2 y+x}{x}\right|=-\log | x |+\log | c |$
$\therefore \frac{1}{2} \log \left|y^2\right|-\frac{1}{2} \log \left|\frac{2 y+x}{x}\right|=\log | c |$
$\therefore \frac{1}{2} \log \left|\frac{y^2}{\frac{2 y+x}{x}}\right|=\log | c | $
$\therefore \frac{1}{2} \log \left|\frac{x y^2}{2 y+x}\right|=\log | c |$

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