The general solution of the differential equa | vedclass

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

Question

(B) Given: $\sec y + (x - e^{\sin y}) \frac{dy}{dx} = 0$Rearranging the equation: $(e^{\sin y} - x) \frac{dy}{dx} = \sec y$$\frac{dx}{dy} = \frac{e^{\

Vedclass · Accepted Answer

$x e^{\sin y} = \frac{e^{2 \sin y}}{2} + c$

Vedclass · Answer

(B) Given: $\sec y + (x - e^{\sin y}) \frac{dy}{dx} = 0$Rearranging the equation: $(e^{\sin y} - x) \frac{dy}{dx} = \sec y$$\frac{dx}{dy} = \frac{e^{\sin y} - x}{\sec y} = (e^{\sin y} - x) \cos y$$\frac{dx}{dy} + x \cos y = e^{\sin y} \cos y$This is a linear differential equation of the form $\frac{dx}{dy} + Px = Q$, where $P = \cos y$ and $Q = e^{\sin y} \cos y$.Integrating factor $IF = e^{\int P dy} = e^{\int \cos y dy} = e^{\sin y}$.The solution is $x \cdot IF = \int (Q \cdot IF) dy + c$.$x e^{\sin y} = \int (e^{\sin y} \cos y \cdot e^{\sin y}) dy + c$.Let $u = \sin y$, then $du = \cos y dy$.$x e^{\sin y} = \int e^{2u} du + c = \frac{e^{2u}}{2} + c$.$x e^{\sin y} = \frac{e^{2 \sin y}}{2} + c$.

The general solution of the differential equation $\sec y + (x - e^{\sin y}) \frac{dy}{dx} = 0$ is...

Similar Questions

The integrating factor of the differential equation $\cos x \frac{dy}{dx} + y \sin x = 1$ is

The solution of the equation $x\frac{dy}{dx} + 3y = x$ is

The general solution of the differential equation $\frac{d y}{d x}=\frac{2 y^2+1}{2 y^3-4 x y+y}$ is

The integrating factor of the differential equation $\frac{dy}{dx} + y \tan x = \sec x$ is . . . . . . .

If $\cos x \frac{dy}{dx} - y \sin x = 6x$ for $0 < x < \frac{\pi}{2}$ and $y(\frac{\pi}{3}) = 0$,then $y(\frac{\pi}{6})$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module