The solution of frac{dy}{dx} = sin(x + y) + c | vedclass

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

Question

(B) Let $v = x + y$. Then $\frac{dv}{dx} = 1 + \frac{dy}{dx}$,so $\frac{dy}{dx} = \frac{dv}{dx} - 1$.Substituting this into the given equation: $\frac

Vedclass · Accepted Answer

$\log \left[ 1 + 	an \left( \frac{x + y}{2} ight) ight] = x + c$

Vedclass · Answer

(B) Let $v = x + y$. Then $\frac{dv}{dx} = 1 + \frac{dy}{dx}$,so $\frac{dy}{dx} = \frac{dv}{dx} - 1$.Substituting this into the given equation: $\frac{dv}{dx} - 1 = \sin v + \cos v$.$\frac{dv}{dx} = 1 + \cos v + \sin v$.Using trigonometric identities: $1 + \cos v = 2\cos^2(v/2)$ and $\sin v = 2\sin(v/2)\cos(v/2)$.$\frac{dv}{dx} = 2\cos^2(v/2) + 2\sin(v/2)\cos(v/2) = 2\cos^2(v/2) [1 + 	an(v/2)]$.Separating variables: $\frac{\sec^2(v/2) dv}{2[1 + 	an(v/2)]} = dx$.Let $u = 1 + 	an(v/2)$,then $du = \frac{1}{2}\sec^2(v/2) dv$.The integral becomes $\int \frac{du}{u} = \int dx$.$\log|u| = x + c$.Substituting back $u = 1 + 	an(v/2)$ and $v = x + y$: $\log \left[ 1 + 	an \left( \frac{x + y}{2} ight) ight] = x + c$.

The solution of $\frac{dy}{dx} = \sin(x + y) + \cos(x + y)$ is

Similar Questions

The solution of the differential equation $(2x - 3y + 5)dx + (9y - 6x - 7)dy = 0$ is

The solution of the differential equation $\sin ^{-1}\left(\frac{dy}{d x}\right)=x+y$ is

If the solution of $\frac{dy}{dx} = (3x + y + 4)^2$ is $\frac{1}{\sqrt{3}} \tan^{-1}(f(x, y)) - x = k$, then $f(1, 2) = $

The particular solution of the differential equation $(1+y^2) dx - xy dy = 0$ with the condition $y(1) = 0$ represents:

The solution of the equation $\frac{dy}{dx} + \sqrt{\frac{1 - y^2}{1 - x^2}} = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module