The solution of the differential equation xfr | vedclass

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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

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