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(B) Given the differential equation: $y - x\frac{dy}{dx} = a(y^2 + \frac{dy}{dx})$Rearranging the terms to group $\frac{dy}{dx}$:$y - ay^2 = x\frac{dy

Vedclass · Accepted Answer

$y = c(x + a)(1 - ay)$

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(B) Given the differential equation: $y - x\frac{dy}{dx} = a(y^2 + \frac{dy}{dx})$Rearranging the terms to group $\frac{dy}{dx}$:$y - ay^2 = x\frac{dy}{dx} + a\frac{dy}{dx}$$y(1 - ay) = (x + a)\frac{dy}{dx}$Separating the variables:$\frac{dy}{y(1 - ay)} = \frac{dx}{x + a}$Using partial fractions for the left side:$\frac{1}{y(1 - ay)} = \frac{1}{y} + \frac{a}{1 - ay}$Integrating both sides:$\int (\frac{1}{y} + \frac{a}{1 - ay}) dy = \int \frac{1}{x + a} dx$$\ln|y| - \ln|1 - ay| = \ln|x + a| + \ln|c|$$\ln|\frac{y}{1 - ay}| = \ln|c(x + a)|$Taking the exponential of both sides:$\frac{y}{1 - ay} = c(x + a)$$y = c(x + a)(1 - ay)$

The solution of the differential equation $y - x\frac{dy}{dx} = a\left( y^2 + \frac{dy}{dx} \right)$ is

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