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

If $y=y(x)$ is the solution of the differential equation $\left(\frac{5+e^x}{2+y}\right) \frac{dy}{dx}+e^x=0$ satisfying $y(0)=1$,then a value of $y(\log 13)$ is

The general solution of the differential equation $\frac{dy}{dx} = \frac{xy+x-2y-2}{xy-2x+y-2}$ is

Let $f(x)$ be a real differentiable function such that $f(0)=1$ and $f(x+y)=f(x)f'(y)+f'(x)f(y)$ for all $x, y \in \mathbb{R}$. Then $\sum_{n=1}^{100} \log_{e} f(n)$ is equal to:

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

Which of the following is true for $y(x)$ that satisfies the differential equation $\frac{dy}{dx} = xy - 1 + x - y$ with the initial condition $y(0) = 0$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module