Answer to Question #207419 in Differential Equations for Tushar

Question

y = 2px + py^2

Answer 1

given Differential equation: y = 2px + py²

where p = $\frac{dy}{dx}$

by putting this we get, y = 2$\frac{dy}{dx}$x + $\frac{dy}{dx}$y²

$\implies$ y = $\frac{dy}{dx}$(2x+y²)

$\implies$ $\frac{dx}{dy}$ y = (2x+y²)

dividing by y both sides, we get

$\implies$ $\frac{dx}{dy}$ = $\frac{2x}{y}$ + y

$\implies$ $\frac{dx}{dy}$ $-$ $\frac{2x}{y}$ = y

now it is a linear differential equation,

integrating factor (I.F) = $e^{\int \frac{-2}{y} dy}$ = $e^{-2\int \frac{1}{y} dy}$ = $e^{-2ln(y)}$ = $e^{ln(\frac{1}{y^2})}$ = $\frac{1}{y^2}$

therefore, the solution of given differential equation is

$\implies$ x$\cdot$ $\frac{1}{y^2}$ = $\int \frac{1}{y^2} \cdot y \cdot dy$ + C where C is any constant

$\implies$ $\frac{x}{y^2}$ = $\int \frac{1}{y} \cdot dy$ + C

$\implies$ $\frac{x}{y^2}$ = $ln(y)$ + C

$\implies$ x = y²ln(y) + Cy²