The general solution of y^2 dx + (x^2 - xy + | vedclass

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

Question

(A) Given differential equation is $y^2 dx + (x^2 - xy + y^2) dy = 0$.Rearranging,we get $\frac{dy}{dx} = \frac{-y^2}{x^2 - xy + y^2}$.This is a homog

Vedclass · Accepted Answer

$	an^{-1}(\frac{y}{x}) = \log y + C$

Vedclass · Answer

(A) Given differential equation is $y^2 dx + (x^2 - xy + y^2) dy = 0$.Rearranging,we get $\frac{dy}{dx} = \frac{-y^2}{x^2 - xy + y^2}$.This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substituting these into the equation: $v + x \frac{dv}{dx} = \frac{-(vx)^2}{x^2 - x(vx) + (vx)^2} = \frac{-v^2}{1 - v + v^2}$.Then $x \frac{dv}{dx} = \frac{-v^2}{1 - v + v^2} - v = \frac{-v^2 - v + v^2 - v^3}{1 - v + v^2} = \frac{-(v^3 + v)}{v^2 - v + 1}$.Separating variables: $\frac{v^2 - v + 1}{v^3 + v} dv = -\frac{1}{x} dx$.Using partial fractions: $\frac{v^2 - v + 1}{v(v^2 + 1)} = \frac{A}{v} + \frac{Bv + C}{v^2 + 1}$. Solving gives $A=1, B=0, C=-1$.So,$(\frac{1}{v} - \frac{1}{v^2 + 1}) dv = -\frac{1}{x} dx$.Integrating both sides: $\log|v| - 	an^{-1}(v) = -\log|x| + C$.$\log|vx| = 	an^{-1}(v) + C$.Since $y = vx$,we have $\log|y| = 	an^{-1}(\frac{y}{x}) + C$,or $	an^{-1}(\frac{y}{x}) = \log y + C$.

The general solution of $y^2 dx + (x^2 - xy + y^2) dy = 0$ is:

Similar Questions

The general solution of the differential equation $x^2 y dx - (x^3 + y^3) dy = 0$ is

The solution of the differential equation $\frac{dy}{dx} = \frac{x + y}{x}$ satisfying the condition $y(1) = 1$ is:

The general solution of $\left(1+e^{\frac{x}{y}}\right) d x+e^{\frac{x}{y}}\left(1-\frac{x}{y}\right) d y=0$ is

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

The general solution of the differential equation $\frac{dy}{dx} + \frac{x+y+1}{x-3y+5} = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module