Question

13. x dy - y dx + a(x2 + y2)dx = 0​

Answer 1

Step-by-step explanation:

We have,

$xdy - ydx + a( {x}^{2} + {y}^{2} )dx = 0$

$\implies \: a( {x}^{2} + {y}^{2} )dx = ydx - xdy$

$\implies \: a \frac{( {x}^{2} + {y}^{2} )}{ {y}^{2} }dx = \frac{ ydx - xdy}{ {y}^{2} }$

$\implies \: a( \frac{ {x}^{2} }{ {y}^{2} } + 1)dx = d( \frac{x}{y} )$

$\implies \: adx = \frac{d( \frac{x}{y} )}{1 + \frac{ {x}^{2} }{ {y}^{2} } }$

$\implies \: a \int \: dx = \int \frac{d( \frac{x}{y} )}{1 + \frac{ {x}^{2} }{ {y}^{2} } }$

$\implies \: ax + c = \tan ^{ - 1} ( \frac{x}{ y} )$

$\implies \tan(ax + c) = \frac{x}{y}$

$\implies \: y = x \cot(ax + c)$