Question

Solve by using Homogeneous Differential equation

Answer 1

Answer:

Let y = vx.

According to the question,

→ ( x - √xy ). dy = y. dx

Transposing 'dx' to the left side and ( x - √xy ) to the right side we get:

$\implies \dfrac{dy}{dx} = \dfrac{y}{(x - \sqrt{xy} )}$

Substituting y = vx, we get:

$\implies \dfrac{dy}{dx} = v + x.\dfrac{dv}{dx}\\\\\\\implies v + x.\dfrac{dv}{dx} = \dfrac{vx}{x - \sqrt{x.vx}}\\\\\\\implies v + x.\dfrac{dv}{dx} = \dfrac{vx}{x - \sqrt{vx^2}}\\\\\\\implies v + x.\dfrac{dv}{dx} = \dfrac{vx}{x - x\sqrt{v}}\\\\\\\implies v + x.\dfrac{dv}{dx} = \dfrac{vx}{x ( 1 - \sqrt{ v})}\\\\\\\implies x.\dfrac{dv}{dx} = \dfrac{v}{( 1 - \sqrt{ v})} - v\\\\\\\implies x.\dfrac{dv}{dx} = \dfrac{v}{( 1 - \sqrt{ v})} - v \times \dfrac{( 1 - \sqrt{ v})}{( 1 - \sqrt{ v})}$

$\implies x. \dfrac{dv}{dx} = \dfrac{ v ( 1 - 1 + \sqrt{v} ) }{ ( 1 - \sqrt{v} ) }\\\\\\\text{Transposing 'v' terms and 'x' terms to LHS and RHS we get: }\\\\\\\implies \dfrac{(1-\sqrt{v})}{v\sqrt{v}}.dv = \dfrac{dx}{x}\\\\\\\implies \dfrac{1}{v\sqrt{v}}.dv - \dfrac{\sqrt{v}}{v\sqrt{v}}.dv = \dfrac{dx}{x}\\\\\\\text{Integrating both sides we get:}\\\\\\\implies \int{v^{-3/2}} \, dv - \int{\dfrac{1}{v}} \, dv = \int{\dfrac{1}{x}} \, dx$

$\implies \boxed{ \bf{ \dfrac{-2}{\sqrt{x}} - ln\:v = ln\: x + C}}\\\\\\\text{Substituting y = vx back we get:}\\\\\\\implies \dfrac{-2}{\sqrt{x}} - ln\:y + ln\:x = ln\: x + C\\\\\\\implies ln\:y = \dfrac{2}{\sqrt{x}} + C\\\\\\\implies \boxed{ \bf{ y = e^{(2/ \sqrt{x}) + C}}}$