Answer to Question #299428 in Differential Equations for BSABE

Question

1. y²dx-(xy+y²)dy=0

Let substitution v=y/x ; y=vx ; dy= vdx+xdv

2. x²dy+(2xy-y²)dx=0

Let substitution v=y/x ; y=vx ; dy= vdx+xdv

Answer 1

1. The given equation can be written as $\dfrac{dy}{dx} = \dfrac{y^2}{xy + y^2}$

Put $y=vx, ~\text{then~} \dfrac{dy}{dx}= v + x\dfrac{dv}{dx}$ . The given equation becomes

$\begin{aligned} v + x\dfrac{dv}{dx} &= \dfrac{v^{2}x^2}{vx^2+v^2x^2}\\ &=\dfrac{v^2}{v+v^2}\\ x\dfrac{dv}{dx} &= \dfrac{v}{1+v} -1\\ &=-\dfrac{v^2}{1+v}\\ \dfrac{1+v}{v^2}dv & = -\dfrac{dx}{x}\\ \dfrac{1}{v^2}dv + \dfrac{1}{v}dv & = -\dfrac{dx}{x}\\ \end{aligned}\\ \text{Integrating both sides, we get}\\ \begin{aligned} -\dfrac{1}{v} + \log v &= -\log x + \log c\\ -\dfrac{1}{v} &= - \log v -\log x + \log c\\ -\dfrac{1}{v} &= - (\log v + \log x - \log c)\\ -\dfrac{1}{v} &= - \log\left(\frac{vx}{c}\right)\\ \dfrac{x}{y} &= \log\left(\frac{y}{c}\right)~~~(\text{Using $y = vx, v = \frac{y}{x}$})\\ e^{\frac{x}{y}} &= \frac{y}{c}\\ \therefore y &= ce^{\frac{x}{y}} \end{aligned}$

2. The equation can be written as $\dfrac{dy}{dx} = \dfrac{y^2 - 2xy}{x^2}$

Put $y=vx, ~\text{then~} \dfrac{dy}{dx}= v + x\dfrac{dv}{dx}$ . The given equation becomes

$\begin{aligned} v + x\dfrac{dv}{dx} &= \dfrac{v^{2}x^2-2vx^2}{x^2}\\ &=v^2 - 2v\\ x\dfrac{dv}{dx} &= v^2 - 3v\\ \dfrac{dv}{v(v-3)} & = \dfrac{dx}{x}\\ \end{aligned}\\ \text{Using partial fractions for the term on LHS, we get}\\ \dfrac{1}{3}\left[\dfrac{dv}{v-3} - \dfrac{dv}{v}\right] = \dfrac{dx}{x}\\ \dfrac{dv}{v-3} - \dfrac{dv}{v} = 3\dfrac{dx}{x}\\ \text{Integrating both sides, we get}\\ \begin{aligned} \log (v-3) - \log v &= 3\log x + \log c\\ \log\left(\dfrac{v-3}{v}\right)&=\log x^3 + \log c\\ \log\left(\dfrac{v-3}{v}\right)&=\log (cx^3)\\ \dfrac{v-3}{v} &=cx^3~~~(\text{taking anti-logarithms})\\ 1 - \frac{3}{v} &=cx^3\\ 1 - \frac{3x}{y} &=cx^3~~~(\text{Using $y = vx$})\\ \frac{3x}{y} &= 1 - cx^3\\ \therefore y &= \frac{3x}{1 - cx^3} \end{aligned}$