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: $(x^2+xy)y'=y^2$.Dividing by $(x^2+xy)$, we get: $\frac{dy}{dx} = \frac{y^2}{x^2+xy} = \frac{y^2}{x(x+y)}$.This i

Vedclass · Accepted Answer

$cy=e^{-\frac{y}{x}}$

Vedclass · Answer

(B) Given the differential equation: $(x^2+xy)y'=y^2$.Dividing by $(x^2+xy)$, we get: $\frac{dy}{dx} = \frac{y^2}{x^2+xy} = \frac{y^2}{x(x+y)}$.This is a homogeneous differential equation. Let $y=vx$, then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.Substituting these into the equation: $v + x\frac{dv}{dx} = \frac{(vx)^2}{x^2+x(vx)} = \frac{v^2x^2}{x^2(1+v)} = \frac{v^2}{1+v}$.Rearranging the terms: $x\frac{dv}{dx} = \frac{v^2}{1+v} - v = \frac{v^2 - v - v^2}{1+v} = \frac{-v}{1+v}$.Separating the variables: $\frac{1+v}{v} dv = -\frac{1}{x} dx$.Integrating both sides: $\int (\frac{1}{v} + 1) dv = -\int \frac{1}{x} dx$.$\ln|v| + v = -\ln|x| + C$.Substituting $v = \frac{y}{x}$: $\ln|\frac{y}{x}| + \frac{y}{x} = -\ln|x| + C$.$\ln|y| - \ln|x| + \frac{y}{x} = -\ln|x| + C$.$\ln|y| + \frac{y}{x} = C$.Taking the exponential of both sides: $e^{\ln|y| + \frac{y}{x}} = e^C$.$y \cdot e^{\frac{y}{x}} = K$ (where $K = e^C$).Rearranging gives $e^{-\frac{y}{x}} = cy$ (where $c = 1/K$).

The general solution of the differential equation $(x^2+xy)y'=y^2$ is

Similar Questions

Which of the following functions are homogeneous?

The substitution $y = z^{\alpha}$ transforms the differential equation $(x^2y^2 - 1)dy + 2xy^3dx = 0$ into a homogeneous differential equation for

The solution of the differential equation $x \frac{dy}{dx} = y - x \tan \left(\frac{y}{x}\right)$ is (Here,$k$ is an arbitrary constant)

The differential equation $y^{\prime} = \frac{y}{x + \sqrt{xy}}$ has a general solution given by (where $C$ is a constant of integration):

The curve satisfying the differential equation $(x^2 - y^2) \, dx + 2xy \, dy = 0$ and passing through the point $(1, 1)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module