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Question

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

Vedclass · Accepted Answer

$(x + a)(1 - ay) = cy$

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

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

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