The general solution of the differential equa | vedclass

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

Question

(D) Given the differential equation: $\frac{dy}{dx} = 1 + x + y + xy$ Factorizing the right side: $\frac{dy}{dx} = (1 + x)(1 + y)$ Separating the vari

Vedclass · Accepted Answer

$y = k e^{x + \frac{x^2}{2}} - 1$

Vedclass · Answer

(D) Given the differential equation: $\frac{dy}{dx} = 1 + x + y + xy$ Factorizing the right side: $\frac{dy}{dx} = (1 + x)(1 + y)$ Separating the variables: $\frac{dy}{1 + y} = (1 + x) dx$ Integrating both sides: $\int \frac{dy}{1 + y} = \int (1 + x) dx$ This gives: $\log(1 + y) = x + \frac{x^2}{2} + c$ Taking the exponential of both sides: $1 + y = e^{x + \frac{x^2}{2} + c}$ $1 + y = e^c \cdot e^{x + \frac{x^2}{2}}$ Let $k = e^c$, then $1 + y = k e^{x + \frac{x^2}{2}}$ Therefore, the general solution is: $y = k e^{x + \frac{x^2}{2}} - 1$

The general solution of the differential equation $\frac{dy}{dx} = 1 + x + y + xy$ is

Similar Questions

The general solution of the differential equation $\frac{dy}{dx} + \frac{1}{\sqrt{1-x^2}} = 0$ is

The solution of the differential equation $e^{2y} (1 + \ln x)dx + \csc y (2 + \cot y)dy = 0$ satisfying $y(1) = \frac{\pi}{2}$ is

The solution of the equation $\frac{dy}{dx} = \frac{y^2 - y - 2}{x^2 + 2x - 3}$ is

The general solution of the differential equation $\frac{dy}{dx} = \frac{3e^{2x} + 3e^{4x}}{e^x + e^{-x}}$ is

The general solution of the differential equation $\frac{dx}{dt} = \frac{x \log x}{t}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module