The solution of cos (x + y) , dy = dx is | vedclass

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

Question

(A) Given the differential equation: $\cos (x + y) \, dy = dx$ ... $(i)$ Let $x + y = v$. Differentiating with respect to $x$,we get $1 + \frac{dy}{dx

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given the differential equation: $\cos (x + y) \, dy = dx$ ... $(i)$ Let $x + y = v$. Differentiating with respect to $x$,we get $1 + \frac{dy}{dx} = \frac{dv}{dx}$,which implies $\frac{dy}{dx} = \frac{dv}{dx} - 1$. Substituting these into $(i)$,we have $\cos v \left( \frac{dv}{dx} - 1 ight) = 1$. This simplifies to $\cos v \frac{dv}{dx} = 1 + \cos v$,or $\frac{\cos v}{1 + \cos v} \, dv = dx$. Using the identity $\cos v = 2 \cos^2(v/2) - 1$,we get $\frac{2 \cos^2(v/2) - 1}{2 \cos^2(v/2)} \, dv = dx$,which simplifies to $\left( 1 - \frac{1}{2} \sec^2(v/2) ight) \, dv = dx$. Integrating both sides,we get $v - 	an(v/2) = x + c$. Substituting $v = x + y$ back,we get $(x + y) - 	an \left( \frac{x + y}{2} ight) = x + c$,which simplifies to $y = 	an \left( \frac{x + y}{2} ight) + c$.

The solution of $\cos (x + y) \, dy = dx$ is

Similar Questions

The general solution of $\frac{dy}{dx} + \sin \left(\frac{x+y}{2}\right) = \sin \left(\frac{x-y}{2}\right)$ is

Find a particular solution of the differential equation $(x+1) \frac{dy}{dx} = 2e^{-y} - 1$,given that $y = 0$ when $x = 0$.

The solution of $x^2 + y^2 \frac{dy}{dx} = 4$ is

The solution of the differential equation $dy - \sin x \sin y \, dx = 0$ is

Let the solution curve of the differential equation $x dy - y dx = \sqrt{x^{2} + y^{2}} dx$,where $x > 0$ and $y(1) = 0$,be $y = y(x)$. Then $y(3)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module