Question

(x^2+y^2-2y)dy-2xdx=0​

Answer 1

Step-by-step explanation:

We have,

$( {x}^{2} + {y}^{2} - 2y)dy - 2xdx = 0$

$\implies {x}^{2} dy + {y}^{2} dy - 2ydy - 2xdx = 0$

$\implies( {x}^{2} + {y}^{2} )dy = 2(ydy + xdx)$

$\implies \sqrt{( { x }^{2} + {y}^{2}) } dy = 2 \frac{(ydy + xdx)}{ \sqrt{ {x}^{2} + {y}^{2} } }$

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

Integrating both sides,

$\implies \int dy = \int \frac{2}{ \sqrt{ {x}^{2} + {y}^{2} } } .d( \sqrt{ {x}^{2} + {y}^{2} })$

$\implies \: y = 2 ln( \sqrt{ {x}^{2} + {y}^{2} } ) + c$

$\implies \: y = ln( {x}^{2} + {y}^{2} ) + ln(c) ...(where \: \: c > 0 \: \: is \: \: a \: \: constant)$

$\implies \: y = ln(c( {x}^{2} + {y}^{2}) )$

$\implies \: c( {x}^{2} + {y}^{2}) = {e}^{y}$

Answer 2

Answer:

Step-by-Step-by-stepstep explanation: