The general solution of the differential equa | vedclass

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

Question

(A) Given the differential equation: $\sec(x-y+1) dy = dx$. Let $v = x-y+1$. Then,$\frac{dv}{dx} = 1 - \frac{dy}{dx}$,which implies $\frac{dy}{dx} = 1

Vedclass · Accepted Answer

$x + \cot \left(\frac{x-y+1}{2}\right) = c$

Vedclass · Answer

(A) Given the differential equation: $\sec(x-y+1) dy = dx$. Let $v = x-y+1$. Then,$\frac{dv}{dx} = 1 - \frac{dy}{dx}$,which implies $\frac{dy}{dx} = 1 - \frac{dv}{dx}$. The equation can be rewritten as $\frac{dy}{dx} = \cos(x-y+1)$. Substituting $v$,we get $1 - \frac{dv}{dx} = \cos(v)$. Rearranging the terms: $\frac{dv}{dx} = 1 - \cos(v)$. Separating the variables: $\frac{dv}{1 - \cos(v)} = dx$. Using the identity $1 - \cos(v) = 2\sin^2(\frac{v}{2})$,we have $\frac{dv}{2\sin^2(\frac{v}{2})} = dx$. This simplifies to $\frac{1}{2} \csc^2(\frac{v}{2}) dv = dx$. Integrating both sides: $\int \frac{1}{2} \csc^2(\frac{v}{2}) dv = \int dx$. $-\cot(\frac{v}{2}) = x + c$. Substituting $v = x-y+1$ back: $-\cot(\frac{x-y+1}{2}) = x + c$,which can be written as $x + \cot(\frac{x-y+1}{2}) = c'$ (where $c' = -c$). Thus,the correct option is $A$.

The general solution of the differential equation $\sec(x-y+1) dy = dx$ is

Similar Questions

Find the general solution of the differential equation: $\frac{dy}{dx} + y = 1$ $(y \neq 1)$.

Let $\alpha |x| = |y| e^{xy-\beta}$,where $\alpha, \beta \in \mathbb{N}$,be the solution of the differential equation $x dy - y dx + xy(x dy + y dx) = 0$ with the initial condition $y(1) = 2$. Then $\alpha + \beta$ is equal to:

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

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

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module