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(A) Given the differential equation $\frac{dy}{dx} = \frac{1 - 2(2x + y)}{1 + (2x + y)}$.Let $v = 2x + y$. Then $\frac{dv}{dx} = 2 + \frac{dy}{dx}$,so

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$4x^2 + 4xy + y^2 - 2x - 2y + c = 0$

Vedclass · Answer

(A) Given the differential equation $\frac{dy}{dx} = \frac{1 - 2(2x + y)}{1 + (2x + y)}$.Let $v = 2x + y$. Then $\frac{dv}{dx} = 2 + \frac{dy}{dx}$,so $\frac{dy}{dx} = \frac{dv}{dx} - 2$.Substituting these into the equation: $\frac{dv}{dx} - 2 = \frac{1 - 2v}{1 + v}$.$\frac{dv}{dx} = \frac{1 - 2v}{1 + v} + 2 = \frac{1 - 2v + 2 + 2v}{1 + v} = \frac{3}{1 + v}$.Separating variables: $(1 + v) dv = 3 dx$.Integrating both sides: $\int (1 + v) dv = \int 3 dx$.$v + \frac{v^2}{2} = 3x + c_1 \Rightarrow 2v + v^2 = 6x + 2c_1$.Substitute $v = 2x + y$ back: $2(2x + y) + (2x + y)^2 = 6x + C$.$4x + 2y + 4x^2 + 4xy + y^2 = 6x + C$.$4x^2 + 4xy + y^2 - 2x + 2y + C = 0$.

The solution of the differential equation $\frac{dy}{dx} = \frac{1 - 2y - 4x}{1 + y + 2x}$ is

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