The solution of the differential equation y - | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given the differential equation: $y - x\frac{dy}{dx} = a(y^2 + \frac{dy}{dx})$Rearranging the terms to separate variables:$y - ay^2 = a\frac{dy}{d

Vedclass · Accepted Answer

$(x + a)(1 - ay) = cy$

Vedclass · Answer

(B) Given the differential equation: $y - x\frac{dy}{dx} = a(y^2 + \frac{dy}{dx})$Rearranging the terms to separate variables:$y - ay^2 = a\frac{dy}{dx} + x\frac{dy}{dx}$$y(1 - ay) = (a + x)\frac{dy}{dx}$Separating the variables:$\frac{dx}{a + x} = \frac{dy}{y(1 - ay)}$Using partial fractions for the right side:$\frac{1}{y(1 - ay)} = \frac{1}{y} + \frac{a}{1 - ay}$Integrating both sides:$\int \frac{dx}{a + x} = \int \left(\frac{1}{y} + \frac{a}{1 - ay}\right) dy$$\ln|a + x| = \ln|y| - \ln|1 - ay| + \ln|c|$$\ln|a + x| = \ln|\frac{cy}{1 - ay}|$Taking the exponential of both sides:$a + x = \frac{cy}{1 - ay}$$(x + a)(1 - ay) = cy$

The solution of the differential equation $y - x\frac{dy}{dx} = a\left(y^2 + \frac{dy}{dx}\right)$ is

Similar Questions

The solution of the differential equation $x\frac{dy}{dx} + y = y^2$ is

The solution of the differential equation $\frac{dy}{dx} = 2xy$ is

Find the general solution of the differential equation: $\sec^{2} x \tan y \, dx + \sec^{2} y \tan x \, dy = 0$

The solution of the differential equation $x dy - y dx = 0$ represents:

The general solution of the differential equation $\frac{d y}{d x}+\frac{\sin (2 x+y)}{\cos x}+2=0$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module