Question

2x^2 dy=(x^2+y^2)dx




solve with homogeneous​

Answer 1

GIVEN :–

• Differentiate equation –

$\\ \implies \bf \: 2 {x}^{2}dy = ( {x}^{2} + {y}^{2} )dx$

TO FIND :–

• Solution of differential equation = ?

SOLUTION :–

$\\ \implies \bf \: 2 {x}^{2}dy = ( {x}^{2} + {y}^{2} )dx$

$\\ \implies \bf \: \dfrac{dy}{dx} = \dfrac{{x}^{2} + {y}^{2}}{2 {x}^{2}} \: \: \: \: \: \: \: \: \: - - - - eq.(1)$

• It's a homogeneous differential equation .

• Now Let's put y = vx –

• So that –

$\\ \implies \bf \: \dfrac{dy}{dx} =v + x \dfrac{dv}{dx}$

• Put the value of dy/dx in eq.(1) –

$\\ \implies \bf \: v + x \dfrac{dv}{dx} =\dfrac{{x}^{2} + {(vx)}^{2}}{2 {x}^{2}}$

$\\ \implies \bf \: v + x \dfrac{dv}{dx} =\dfrac{{x}^{2} +{v}^{2} {x}^{2} }{2 {x}^{2}}$

$\\ \implies \bf \: v + x \dfrac{dv}{dx} =\dfrac{{x}^{2}(1 +{v}^{2})}{2 {x}^{2}}$

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

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

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

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

$\\ \implies \bf \:\dfrac{2dv}{{(v - 1)}^{2}} =\dfrac{dx}{x}$

• Integration on both sides –

$\\ \implies \bf \: \int\dfrac{2dv}{{(v - 1)}^{2}} = \int\dfrac{dx}{x}$

• Put v - 1 = t :–

• And –

$\\ \implies \bf \: \dfrac{dv}{dt} = 1$

$\\ \implies \bf \:dv=dt$

• So that –

$\\ \implies \bf \: \int\dfrac{2dt}{{(t)}^{2}} = \int\dfrac{dx}{x}$

$\\ \implies \bf \: - \dfrac{2}{t} = log(x) + c$

• Replace 't' –

$\\ \implies \bf \: - \dfrac{2}{v - 1} = log(x) + c$

• Now replace 'v' –

$\\ \implies \bf \: \dfrac{2}{1 - \dfrac{y}{x} } = log(x) + c$

$\\ \large\implies{ \boxed{ \bf \dfrac{2x}{x -y} = log(x) + c}}$