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(B) Given the differential equation $x + y\frac{dy}{dx} = 2y$.Rearranging the terms,we get $y\frac{dy}{dx} = 2y - x$,which implies $\frac{dy}{dx} = \f

Vedclass · Accepted Answer

$\log (y - x) = c + \frac{x}{y - x}$

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(B) Given the differential equation $x + y\frac{dy}{dx} = 2y$.Rearranging the terms,we get $y\frac{dy}{dx} = 2y - x$,which implies $\frac{dy}{dx} = \frac{2y - x}{y}$.This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.Substituting these into the equation: $v + x\frac{dv}{dx} = \frac{2vx - x}{vx} = \frac{2v - 1}{v}$.$x\frac{dv}{dx} = \frac{2v - 1}{v} - v = \frac{2v - 1 - v^2}{v} = -\frac{(v - 1)^2}{v}$.Separating the variables: $\frac{v}{(v - 1)^2} dv = -\frac{dx}{x}$.Using partial fractions: $\frac{v}{(v - 1)^2} = \frac{v - 1 + 1}{(v - 1)^2} = \frac{1}{v - 1} + \frac{1}{(v - 1)^2}$.Integrating both sides: $\int \left( \frac{1}{v - 1} + \frac{1}{(v - 1)^2} \right) dv = -\int \frac{dx}{x}$.$\log |v - 1| - \frac{1}{v - 1} = -\log |x| + c$.Since $v = \frac{y}{x}$,we have $v - 1 = \frac{y - x}{x}$.$\log |\frac{y - x}{x}| + \log |x| = \frac{1}{\frac{y - x}{x}} + c$.$\log |y - x| = \frac{x}{y - x} + c$.

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

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