If cos frac{y}{x} = A log x + C is the genera | vedclass

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

Question

(B) Given the differential equation: $(x \sin \frac{y}{x}) dy = (y \sin \frac{y}{x} - x) dx$ Rearranging the terms to find $\frac{dy}{dx}$: $\frac{dy}

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) Given the differential equation: $(x \sin \frac{y}{x}) dy = (y \sin \frac{y}{x} - x) dx$ Rearranging the terms to find $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{y \sin(y/x) - x}{x \sin(y/x)} = \frac{y}{x} - \frac{1}{\sin(y/x)}$ Let $v = \frac{y}{x}$,then $y = vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ Substituting these into the equation: $v + x \frac{dv}{dx} = v - \frac{1}{\sin v}$ $x \frac{dv}{dx} = -\frac{1}{\sin v}$ Separating the variables: $-\sin v \, dv = \frac{1}{x} \, dx$ Integrating both sides: $-\int \sin v \, dv = \int \frac{1}{x} \, dx$ $\cos v = \log x + C$ Substituting $v = \frac{y}{x}$ back: $\cos \frac{y}{x} = \log x + C$ Comparing this with the given solution $\cos \frac{y}{x} = A \log x + C$,we find $A = 1$.

If $\cos \frac{y}{x} = A \log x + C$ is the general solution of $(x \sin \frac{y}{x}) dy = (y \sin \frac{y}{x} - x) dx$,then $A =$

Similar Questions

$A$ bag contains $5$ red balls and $3$ green balls. $A$ ball is selected at random and not replaced. $A$ second ball is then selected. The probability of selecting one red ball and one green ball is

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

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

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

The slope of the tangent at $(x, y)$ to a curve passing through $\left( 1, \frac{\pi}{4} \right)$ is given by $\frac{y}{x} - \cos^2\left( \frac{y}{x} \right)$. Then the equation of the curve is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module