Question

Solve the Homogeneous differential equation xydx - (x² + 3y²) dy = 0​

Answer 1

Step-by-step explanation:

We have,

$\tt{xy \: dx - \left( {x}^{2} + 3 {y}^{2} \right) \: dy = 0}$

$\tt{ \implies \: xy \: dx = \left( {x}^{2} + 3 {y}^{2} \right) \: dy}$

$\tt{ \implies \dfrac{dx }{dy} = \dfrac{ {x}^{2} + 3 {y}^{2}}{xy} }$

Put $\sf{y=vx}$

$\implies\sf{\dfrac{dy}{dx}=v+x\,\dfrac{dv}{dx}}$

$\tt{ \implies \dfrac{dy}{dx} = \dfrac{xy}{ {x}^{2} + 3 {y}^{2}} }$

$\tt{ \implies v + x \dfrac{dv}{dx} = \dfrac{v \: x^{2} }{ {x}^{2} + 3 {v}^{2} \cdot{x}^{2} } }$

$\tt{ \implies v + x \dfrac{dv}{dx} = \dfrac{v }{ 1+ 3 {v}^{2} } }$

$\tt{ \implies x \dfrac{dv}{dx} = \dfrac{v }{ 1+ 3 {v}^{2} } - v}$

$\tt{ \implies x \dfrac{dv}{dx} = \dfrac{v - v - 3 {v}^{3} }{ 1+ 3 {v}^{2} } }$

$\tt{ \implies x \dfrac{dv}{dx} = \dfrac{ - 3 {v}^{3} }{ 1+ 3 {v}^{2} } }$

$\tt{ \implies \dfrac{1 + 3v^{2} }{3 {v}^{3} } \: dv = - \dfrac {dx }{ x } }$

Integrating both sides, we get,

$\displaystyle \tt{ \implies \int\dfrac{1 + 3v^{2} }{3 {v}^{3} } \: dv = - \int \dfrac {dx }{ x } }$

$\displaystyle \tt{ \implies \int\dfrac{1 }{3 {v}^{3} } \: dv + \int \dfrac{3 {v}^{2} }{3 {v}^{3} } \: dv = - \int \dfrac {dx }{ x } }$

$\displaystyle \tt{ \implies \dfrac{1}{3} \int\dfrac{dv}{ {v}^{3} } + \int \dfrac{dv }{v} = - \int \dfrac {dx }{ x } }$

$\displaystyle \tt{ \implies - \dfrac{1}{3} \cdot \dfrac{1}{2 {v}^{2} } + \ln | v | = - \ln | x | + c }$

$\displaystyle \tt{ \implies - \dfrac{1}{6 {v}^{2} } + \ln | v | = - \ln | x | + C }$

$\displaystyle \tt{ \implies - \dfrac{ {x}^{2} }{6 {y}^{2} } + \ln \left | \dfrac{y}{x} \right| = - \ln | x | + C }$

$\displaystyle \tt{ \implies - \dfrac{ {x}^{2} }{6 {y}^{2} } + \ln | y | - \ln |x| = - \ln | x | + C }$

$\displaystyle \tt{ \implies - \dfrac{ {x}^{2} }{6 {y}^{2} } + \ln | y | = C }$

Answer 2

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

Given homogeneous differential equation is

$\rm :\longmapsto\:xydx - ( {x}^{2} + {3y}^{2})dy = 0$

$\rm :\longmapsto\:xydx = ( {x}^{2} + {3y}^{2})dy$

$\rm :\longmapsto\:\dfrac{dx}{dy} = \dfrac{ {x}^{2} + {3y}^{2} }{xy}$

To solve this Homogeneous differential equation, we substitute

$\red{\rm :\longmapsto\:x = vy}$

So, above differential equation reduces to

$\rm :\longmapsto\:\dfrac{d}{dy}(vy) = \dfrac{ {(vy)}^{2} + {3y}^{2} }{(vy)y}$

$\rm :\longmapsto\:y\dfrac{d}{dy} v + v\dfrac{d}{dy} y= \dfrac{ {v}^{2} {y}^{2} + {3y}^{2} }{ {vy}^{2} }$

$\rm :\longmapsto\:y\dfrac{dv}{dy} + v= \dfrac{ {y}^{2} ({v}^{2} + 3) }{ {vy}^{2} }$

$\rm :\longmapsto\:y\dfrac{dv}{dy} + v= \dfrac{{v}^{2} + 3}{v}$

$\rm :\longmapsto\:y\dfrac{dv}{dy}= \dfrac{{v}^{2} + 3}{v} - v$

$\rm :\longmapsto\:y\dfrac{dv}{dy}= \dfrac{{v}^{2}+ 3 - {v}^{2} }{v}$

$\rm :\longmapsto\:y\dfrac{dv}{dy}= \dfrac{3}{v}$

$\rm :\longmapsto\:vdv=3 \dfrac{dy}{y}$

$\rm :\longmapsto\:\displaystyle \int \:vdv=3 \displaystyle \int \:\dfrac{dy}{y}$

$\rm :\longmapsto\:\dfrac{ {v}^{2} }{2} = 3logy + c$

$\rm :\longmapsto\:\dfrac{{x}^{2}}{2 {y}^{2} } = 3logy + c$

$\rm :\longmapsto\:\dfrac{{x}^{2}}{{y}^{2} } = 6logy + 2c$

$\rm :\longmapsto\:\dfrac{{x}^{2}}{{y}^{2} } = 6logy +d \: \: \: (where \: d = 2c)$