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 particular solution of the differential equation $\frac{dy}{dx} - e^x = y e^x$,when $x = 0$ and $y = 1$ is

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

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

If $\lim _{x \rightarrow \infty} y(x)=\frac{\pi}{2}$,then the solution of $x^3 \sin y \frac{d y}{d x}=2$ is $\cos y=$

The general solution of $\left(x \frac{dy}{dx} - y\right) \sin \frac{y}{x} = x^3 e^x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module