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

Find a particular solution satisfying the given condition: $\frac{dy}{dx} + 2y \tan x = \sin x$; $y = 0$ when $x = \frac{\pi}{3}$.

If $x \log x \frac{dy}{dx} + y = \log x^2$ and $y(e) = 0$,then $y(e^2) = $

Find a particular solution of the differential equation $\frac{dy}{dx} + y \cot x = 4x \csc x$ $(x \neq 0),$ given that $y=0$ when $x=\frac{\pi}{2}$.

Let $y=y(x)$ be the solution of the differential equation $\sec x \frac{dy}{dx} - 2y = 2 + 3 \sin x$, where $x \in (-\frac{\pi}{2}, \frac{\pi}{2})$ and $y(0) = -\frac{7}{4}$. Then $y(\frac{\pi}{6})$ is equal to:

The general solution of the differential equation $x^{2} dy - 2xy dx = x^{4} \cos x dx$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module