The general solution of the differential equa | vedclass

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

Question

(C) Given differential equation: $\cos (x+y) dy = dx$ Rearranging, we get: $\frac{dx}{dy} = \cos (x+y)$ Let $x+y = v$. Then, differentiating with resp

Vedclass · Accepted Answer

$y = 	an \left(\frac{x+y}{2}ight) + c$

Vedclass · Answer

(C) Given differential equation: $\cos (x+y) dy = dx$ Rearranging, we get: $\frac{dx}{dy} = \cos (x+y)$ Let $x+y = v$. Then, differentiating with respect to $y$, we get: $\frac{dx}{dy} + 1 = \frac{dv}{dy} \implies \frac{dx}{dy} = \frac{dv}{dy} - 1$ Substituting this into the equation: $\frac{dv}{dy} - 1 = \cos v$ $\frac{dv}{dy} = 1 + \cos v$ Separating variables: $\int \frac{dv}{1 + \cos v} = \int dy$ Using the identity $1 + \cos v = 2 \cos^2 \frac{v}{2}$: $\int \frac{dv}{2 \cos^2 \frac{v}{2}} = \int dy$ $\frac{1}{2} \int \sec^2 \frac{v}{2} dv = \int dy$ Integrating both sides: $	an \frac{v}{2} = y + c$ Substituting $v = x+y$ back: $	an \frac{x+y}{2} = y + c$ Thus, $y = 	an \frac{x+y}{2} + c$.

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

Similar Questions

The particular solution of the differential equation $\frac{dy}{dx} = \frac{x+y+1}{x+y-1}$ when $x = \frac{2}{3}$ and $y = \frac{1}{3}$ is

Let $f: R \rightarrow R$ be a differentiable function with $f(0)=0$. If $y=f(x)$ satisfies the differential equation $\frac{dy}{dx}=(2+5y)(5y-2)$,then the value of $\lim_{x \rightarrow -\infty} f(x)$ is:

If $2 f(x) = f^{\prime}(x)$ and $f(0) = 3$,then the value of $f(2)$ is

General solution of the differential equation $\log \left(\frac{d y}{d x}\right)=a x+b y$ is

For solving $\frac{dy}{dx} = (4x + y + 1)$,the suitable substitution is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module