The solution of the differential equation fra | vedclass

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

Question

(B) Given,$\frac{dy}{dx} = \sin(x+y) 	an(x+y) - 1$. Let $x+y = z$. Then,differentiating with respect to $x$,we get $1 + \frac{dy}{dx} = \frac{dz}{dx}

Vedclass · Accepted Answer

$x + \operatorname{cosec}(x+y) = c$

Vedclass · Answer

(B) Given,$\frac{dy}{dx} = \sin(x+y) 	an(x+y) - 1$. Let $x+y = z$. Then,differentiating with respect to $x$,we get $1 + \frac{dy}{dx} = \frac{dz}{dx}$,so $\frac{dy}{dx} = \frac{dz}{dx} - 1$. Substituting this into the differential equation: $\frac{dz}{dx} - 1 = \sin z 	an z - 1$ $\frac{dz}{dx} = \sin z 	an z = \sin z \cdot \frac{\sin z}{\cos z} = \frac{\sin^2 z}{\cos z}$. Rearranging the variables: $\int \frac{\cos z}{\sin^2 z} dz = \int dx$. Let $\sin z = t$,then $\cos z dz = dt$. $\int \frac{1}{t^2} dt = x + c$ $- \frac{1}{t} = x + c$ $- \operatorname{cosec} z = x + c$ Substituting $z = x+y$ back: $- \operatorname{cosec}(x+y) = x + c$ $x + \operatorname{cosec}(x+y) = c$ (where $c$ is an arbitrary constant).

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

Similar Questions

The solution of the differential equation $\frac{dx}{dy} + 2yx = 2y$ which passes through the point $(2,0)$ is

On solving $\frac{dy}{dx} = \frac{x-y+3}{2x-2y+5}$,the solution obtained is $x = 2(x-y) + \log(t) + c$,find $t$.

The general solution of the differential equation $\frac{1}{x} \frac{dy}{dx} = \tan^{-1} x$ is

$A$ particle starts at the origin and moves along the $x$-axis in such a way that its velocity at the point $(x, 0)$ is given by the formula $\frac{dx}{dt} = \cos^2(\pi x)$. Then the particle never reaches the point on:

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module