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

If $y=y(x)$ satisfies $\left(\frac{2+\sin x}{1+y}\right) \frac{dy}{dx} = -\cos x$ such that $y(0)=2$,then $y\left(\frac{\pi}{2}\right)$ is equal to

The solution of the differential equation $dy - \sin x \sin y \, dx = 0$ is

The solution of $x^2 + y^2 \frac{dy}{dx} = 4$ is

The particular solution of the differential equation $\log\left(\frac{dy}{dx}\right) = x$,when $x = 0, y = 1$ is .....

The general solution of the differential equation $\frac{dy}{dx} = e^{y+x} + e^{y-x}$ is, where $c$ is an arbitrary constant.

Vedclass Test Series

Exam Paper Generator

Online Exam Module