The solution of frac{d y}{d x}=frac{y^2}{x y- | vedclass

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

Question

(B) Given, $\frac{d y}{d x}=\frac{y^2}{x y-x^2}$.This is a homogeneous differential equation. Let $y = vx$, then $\frac{d y}{d x} = v + x \frac{d v}{d

Vedclass · Accepted Answer

$e^{y / x}=k y$

Vedclass · Answer

(B) Given, $\frac{d y}{d x}=\frac{y^2}{x y-x^2}$.This is a homogeneous differential equation. Let $y = vx$, then $\frac{d y}{d x} = v + x \frac{d v}{d x}$.Substituting these into the equation:$v + x \frac{d v}{d x} = \frac{(vx)^2}{x(vx) - x^2} = \frac{v^2 x^2}{x^2(v - 1)} = \frac{v^2}{v - 1}$.$x \frac{d v}{d x} = \frac{v^2}{v - 1} - v = \frac{v^2 - v^2 + v}{v - 1} = \frac{v}{v - 1}$.Separating the variables:$\frac{v - 1}{v} d v = \frac{d x}{x}$.$(1 - \frac{1}{v}) d v = \frac{d x}{x}$.Integrating both sides:$\int (1 - \frac{1}{v}) d v = \int \frac{d x}{x}$.$v - \ln |v| = \ln |x| + C$.Substituting $v = \frac{y}{x}$:$\frac{y}{x} - \ln |\frac{y}{x}| = \ln |x| + C$.$\frac{y}{x} = \ln |\frac{y}{x}| + \ln |x| + C = \ln |y| + C$.$e^{y/x} = e^{\ln |y| + C} = e^C \cdot y = ky$ (where $k = e^C$ is a constant).Thus, $e^{y/x} = ky$.

The solution of $\frac{d y}{d x}=\frac{y^2}{x y-x^2}$ is

Similar Questions

Three fair coins are tossed. If both heads and tails appear,then the probability that exactly one head appears is:

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

Solve the differential equation $(x \,dy-y \,dx) y\, \sin \left(\frac{y}{x}\right)=(y \,dx+x\, dy) x\, \cos \left(\frac{y}{x}\right)$.

If the gradient of the tangent at any point $(x, y)$ of a curve which passes through the point $\left( 1, \frac{\pi}{4} \right)$ is $\left\{ \frac{y}{x} - \sin^2\left( \frac{y}{x} \right) \right\}$,then the equation of the curve is:

If $P(A') = 0.6$,$P(B) = 0.8$,and $P(B/A) = 0.3$,then $P(A/B) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module