The solution of the differential equation fra | vedclass

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

Question

(A) Given the differential equation: $\frac{dy}{dx} = (x+y)^2$ $(i)$ Let $x+y = t$. Differentiating with respect to $x$,we get $1 + \frac{dy}{dx} = \f

Vedclass · Accepted Answer

$	an^{-1}(x+y) = x+C$

Vedclass · Answer

(A) Given the differential equation: $\frac{dy}{dx} = (x+y)^2$ $(i)$ Let $x+y = t$. Differentiating with respect to $x$,we get $1 + \frac{dy}{dx} = \frac{dt}{dx}$,which implies $\frac{dy}{dx} = \frac{dt}{dx} - 1$. Substituting this into equation $(i)$: $\frac{dt}{dx} - 1 = t^2$ $\frac{dt}{dx} = t^2 + 1$ Separating the variables: $\frac{dt}{t^2 + 1} = dx$ Integrating both sides: $\int \frac{dt}{t^2 + 1} = \int dx$ $	an^{-1}(t) = x + C$ Substituting $t = x+y$ back: $	an^{-1}(x+y) = x + C$

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

Similar Questions

The general solution of the equation $({e^y} + 1)\cos x \, dx + {e^y}\sin x \, dy = 0$ is

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

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

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

If $y=y(x)$ and $\frac{2+\sin x}{y+1}\left(\frac{d y}{d x}\right)=-\cos x$,with $y(0)=1$,then $y\left(\frac{\pi}{2}\right)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module