The solution of the equation frac{dy}{dx} = f | vedclass

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(A) Given the differential equation $\frac{dy}{dx} = \frac{x}{2y - x}$.Since this is a homogeneous differential equation,we substitute $y = vx$,which

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$(x - y)(x + 2y)^2 = c$

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(A) Given the differential equation $\frac{dy}{dx} = \frac{x}{2y - x}$.Since this is a homogeneous differential equation,we substitute $y = vx$,which implies $\frac{dy}{dx} = v + x\frac{dv}{dx}$.Substituting these into the equation: $v + x\frac{dv}{dx} = \frac{x}{2vx - x} = \frac{1}{2v - 1}$.Rearranging for $\frac{dv}{dx}$: $x\frac{dv}{dx} = \frac{1}{2v - 1} - v = \frac{1 - v(2v - 1)}{2v - 1} = \frac{1 - 2v^2 + v}{2v - 1} = -\frac{2v^2 - v - 1}{2v - 1} = -\frac{(2v + 1)(v - 1)}{2v - 1}$.Separating the variables: $\frac{2v - 1}{(2v + 1)(v - 1)} dv = -\frac{dx}{x}$.Using partial fractions: $\frac{2v - 1}{(2v + 1)(v - 1)} = \frac{A}{2v + 1} + \frac{B}{v - 1}$.Solving for coefficients,we get $\frac{4/3}{2v + 1} + \frac{1/3}{v - 1} dv = -\frac{dx}{x}$.Integrating both sides: $\frac{2}{3}\ln|2v + 1| + \frac{1}{3}\ln|v - 1| = -\ln|x| + C$.Multiplying by $3$: $2\ln|2v + 1| + \ln|v - 1| = -3\ln|x| + C'$.$\ln|(2v + 1)^2(v - 1)| = \ln|x^{-3}| + C'$.$(2v + 1)^2(v - 1) = \frac{c}{x^3}$.Substituting $v = \frac{y}{x}$: $(2\frac{y}{x} + 1)^2(\frac{y}{x} - 1) = \frac{c}{x^3}$.$(\frac{2y + x}{x})^2(\frac{y - x}{x}) = \frac{c}{x^3}$.$(2y + x)^2(y - x) = c$. Since $(y - x) = -(x - y)$,we get $(x - y)(x + 2y)^2 = c$.

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

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