Question

$x \: dx + y \: dy + ({x}^{2} + {y}^{2} )dx= 0$
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Answer 1

$\large\underline{\sf{Solution-}}$

Given Differential equation is

$\rm :\longmapsto\:xdx + ydy + ( {x}^{2} + {y}^{2})dx = 0$

can be rewritten as

$\rm :\longmapsto\:xdx + ydy = - ( {x}^{2} + {y}^{2})dx$

$\rm :\longmapsto\:\dfrac{xdx \: + \: ydy}{ {x}^{2} + {y}^{2} } = - \: dx$

On multiply both sides by 2,

$\rm :\longmapsto\:\dfrac{2xdx \: + \: 2ydy}{ {x}^{2} + {y}^{2} } = - 2 \: dx$

can be further rewritten as

$\rm :\longmapsto\:\dfrac{d( \: {x}^{2} + \: {y}^{2} )}{ {x}^{2} + {y}^{2} } = - 2 \: dx$

can be further rewritten as

$\rm :\longmapsto\:d \: \bigg( log( {x}^{2} + {y}^{2} ) \bigg) = -2 \: dx$

On integrating both sides, we get

$\rm :\longmapsto\: log( {x}^{2} + {y}^{2}) \: = \: - 2 \: x \: + \: c$

$\rm :\longmapsto\: log( {x}^{2} + {y}^{2}) \: + 2 \: x \: = \: c$

Additional Information :-

$\red{ \boxed{ \sf{ \:d(xy) = ydx \: + \: xdy }}}$

$\red{ \boxed{ \sf{ d( \frac{y}{x} )\: = \: \dfrac{xdy - ydx}{ {x}^{2} } }}}$

$\red{ \boxed{ \sf{ d( \frac{x}{y} )\: = \: \dfrac{ydx - xdy}{ {y}^{2} } }}}$

$\red{ \boxed{ \sf{d \: {tan}^{ - 1} \frac{y}{x} \: = \: \dfrac{xdy - ydx}{ {x}^{2} + {y}^{2} } }}}$

$\red{ \boxed{ \sf{d \: {tan}^{ - 1} \frac{x}{y} \: = \: \dfrac{ydx - xdy}{ {x}^{2} + {y}^{2} } }}}$

$\red{ \boxed{ \sf{d\:log ( {x}^{2}+{y}^{2})\: = \: \dfrac{2xdx + 2ydy}{ {x}^{2} + {y}^{2} } }}}$