Answer to Question #146749 in Differential Equations for Nikhil Singh

Question

Solve the differential equation
(x-y^2)dx+2xy=0

Answer 1

$$\begin{aligned} (x-y^2)dx + 2xy dy &= 0\\ 2y\frac{dy}{dx} - \frac{y^2}{x} &= -1\\ \\ Let\ v = y^2 \\ dv = 2ydy\\ \\ Then, \frac{dv}{dx} - \frac{v}{x} &= -1\ ---(i)\\ \\ P &= \frac{-1}{x}\\ \int{Pdx} &= -\int{\frac{1}{x}dx} = -\ln x \\ &= \ln{\frac{1}{x}}\\ \\ \therefore The\ I.F &= e^{ln\frac{1}{x}} = \frac{1}{x} \end{aligned}$$

Multiplying equation (i) by the I.F, we get,

$\begin{aligned} \frac{1}{x}\frac{dv}{dx} - \frac{1}{x}\frac{v}{x} &= -\frac{1}{x}\\ \frac{1}{x}\frac{dv}{dx} - \frac{v}{x^2} &= -\frac{1}{x} \end{aligned}$

Integrating both sides with respect to x,

$\begin{aligned} \frac{v}{x} &= -\int{\frac{1}{x}dx} + b = -\ln{x} + b\\ \\ \therefore y^2 &=- x\ln{x}+bx \end{aligned}$