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 the differential equation: $\left(x \sin \frac{y}{x}\right) dy = \left(y \sin \frac{y}{x} - x\right) dx$. Rearranging the terms,we get: $\fr

Vedclass · Accepted Answer

$\log |x| - \cos \frac{y}{x} = c$

Vedclass · Answer

(B) Given the differential equation: $\left(x \sin \frac{y}{x}\right) dy = \left(y \sin \frac{y}{x} - x\right) dx$. Rearranging the terms,we get: $\frac{dy}{dx} = \frac{y \sin(y/x) - x}{x \sin(y/x)}$. Since this is a homogeneous differential equation,let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$. Substituting these into the equation: $v + x \frac{dv}{dx} = \frac{vx \sin v - x}{x \sin v} = \frac{v \sin v - 1}{\sin v} = v - \frac{1}{\sin v}$. Subtracting $v$ from both sides: $x \frac{dv}{dx} = -\frac{1}{\sin v}$. Separating the variables: $\sin v \, dv = -\frac{1}{x} dx$. Integrating both sides: $\int \sin v \, dv = -\int \frac{1}{x} dx$. $-\cos v = -\log |x| + C$. Rearranging gives: $\log |x| - \cos v = C$. Substituting $v = \frac{y}{x}$ back,we get: $\log |x| - \cos \frac{y}{x} = C$.

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

Similar Questions

Find the particular solution satisfying the given condition: $2xy + y^2 - 2x^2 \frac{dy}{dx} = 0$; $y = 2$ when $x = 1$.

The integral curve satisfying $y' = \frac{x^2 + y^2}{x^2 - y^2}$ with $y(1) = 2$ has a slope at the point $(1, 0)$ equal to:

For the differential equation $(x^2 + y^2) dy = xy dx$, it is given that $y(1) = 1$ and $y(x_0) = e$. Find the value of $x_0$.

Two aeroplanes $I$ and $II$ bomb a target in succession. The probabilities of $I$ and $II$ scoring a hit are $0.3$ and $0.2$,respectively. The second plane will bomb only if the first misses the target. The probability that the target is hit by the second plane is

If $x \frac{dy}{dx} = y(\log y - \log x + 1)$,then the solution of the equation is

Vedclass Test Series

Exam Paper Generator

Online Exam Module