The solution of the differential equation xfr | vedclass

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

Question

(A) Given the differential equation: $x\frac{dy}{dx} + y = y^2$Rearranging the terms,we get: $x\frac{dy}{dx} = y^2 - y$Separating the variables: $\fra

Vedclass · Accepted Answer

$y = 1 + cxy$

Vedclass · Answer

(A) Given the differential equation: $x\frac{dy}{dx} + y = y^2$Rearranging the terms,we get: $x\frac{dy}{dx} = y^2 - y$Separating the variables: $\frac{dy}{y^2 - y} = \frac{dx}{x}$Using partial fractions: $\left( \frac{1}{y-1} - \frac{1}{y} \right) dy = \frac{dx}{x}$Integrating both sides: $\int \left( \frac{1}{y-1} - \frac{1}{y} \right) dy = \int \frac{dx}{x}$$\log|y-1| - \log|y| = \log|x| + \log|c|$$\log\left| \frac{y-1}{y} \right| = \log|cxy|$Taking the exponential of both sides: $\frac{y-1}{y} = cxy$$y - 1 = cxy^2$ (Note: The standard form derived from the integration is $\frac{y-1}{y} = cx$,which simplifies to $y-1 = cxy$. Thus,$y = 1 + cxy$ is the correct solution form).

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

Similar Questions

Let $x=x(t)$ and $y=y(t)$ be solutions of the differential equations $\frac{dx}{dt}+ax=0$ and $\frac{dy}{dt}+by=0$ respectively,where $a, b \in R$. Given that $x(0)=2$,$y(0)=1$,and $3y(1)=2x(1)$,the value of $t$ for which $x(t)=y(t)$ is:

The solution of the differential equation $\frac{dx}{x} + \frac{dy}{y} = 0$ is

The solution of $\frac{dy}{dx} = (\frac{x}{y})^{-1/3}$ is

The solution of the differential equation $\sec^2 x \tan y \, dx + \sec^2 y \tan x \, dy = 0$ is

The equation of the curve that passes through the point $(1, 2)$ and satisfies the differential equation $\frac{dy}{dx} = \frac{-2xy}{x^2 + 1}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module