The solution of y' = 1 + x + y^2 + xy^2,y(0) | vedclass

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

Question

(D) Given the differential equation $\frac{dy}{dx} = 1 + x + y^2 + xy^2$. Factoring the right side,we get $\frac{dy}{dx} = (1 + x) + y^2(1 + x) = (1 +

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Given the differential equation $\frac{dy}{dx} = 1 + x + y^2 + xy^2$. Factoring the right side,we get $\frac{dy}{dx} = (1 + x) + y^2(1 + x) = (1 + x)(1 + y^2)$. Separating the variables,we have $\frac{dy}{1 + y^2} = (1 + x) dx$. Integrating both sides,$\int \frac{dy}{1 + y^2} = \int (1 + x) dx$. This yields $	an^{-1} y = x + \frac{x^2}{2} + c$. Using the initial condition $y(0) = 0$,we substitute $x = 0$ and $y = 0$: $	an^{-1}(0) = 0 + \frac{0^2}{2} + c$,which gives $c = 0$. Therefore,$	an^{-1} y = x + \frac{x^2}{2}$,which implies $y = 	an \left( x + \frac{x^2}{2} ight)$.

The solution of $y' = 1 + x + y^2 + xy^2$,$y(0) = 0$ is

Similar Questions

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

The solution of $(1+y^2) dx - xy dy = 0$,$y(1)=0$ represents a conic. Its eccentricity is

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

If $\left( \frac{2 + \sin x}{1 + y} \right) \frac{dy}{dx} = - \cos x$ and $y(0) = 1$,then $y\left( \frac{\pi}{2} \right) = $

The solution of the differential equation $\frac{dy}{dx} = (ae^{bx} + c\cos mx)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module