The particular solution of frac{dy}{dx} = 1 + | vedclass

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

Question

(D) Given the differential equation: $\frac{dy}{dx} = 1 + x + y^2 + xy^2$ Factorizing the right side: $\frac{dy}{dx} = (1 + x)(1 + y^2)$ Separating th

Vedclass · Accepted Answer

$y = 	an \left(x + \frac{x^2}{2}ight)$

Vedclass · Answer

(D) Given the differential equation: $\frac{dy}{dx} = 1 + x + y^2 + xy^2$ Factorizing the right side: $\frac{dy}{dx} = (1 + x)(1 + y^2)$ Separating the variables: $\int \frac{dy}{1 + y^2} = \int (1 + x) dx$ Integrating both sides: $	an^{-1}(y) = x + \frac{x^2}{2} + C$ Given the initial condition $y(0) = 0$: $	an^{-1}(0) = 0 + \frac{0^2}{2} + C \Rightarrow C = 0$ Thus,the particular solution is: $	an^{-1}(y) = x + \frac{x^2}{2}$ Therefore: $y = 	an \left(x + \frac{x^2}{2}ight)$

The particular solution of $\frac{dy}{dx} = 1 + x + y^2 + xy^2$,when $y(0) = 0$,is

Similar Questions

Let the solution curve of the differential equation $x dy - y dx = \sqrt{x^{2} + y^{2}} dx$, where $x > 0$ and $y(1) = 0$, be $y = y(x)$. Then $y(3)$ is equal to:

The equation of the curve passing through the point $(0, -2)$ given that at any point $(x, y)$ on the curve,the product of the slope of its tangent and the $y$-coordinate of the point is equal to the $x$-coordinate of the point,is

The particular solution of $\log \left(\frac{dy}{dx}\right) = 3x + 4y$ at $x = y = 0$ is

The solution of $\cos (x + y) \, dy = dx$ is

Find the equation of a curve passing through the point $(0, -2)$ given that at any point $(x, y)$ on the curve,the product of the slope of its tangent and the $y$-coordinate of the point is equal to the $x$-coordinate of the point.

Vedclass Test Series

Exam Paper Generator

Online Exam Module