The general solution of frac{dy}{dx} = frac{x | vedclass

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(C) Given the differential equation: $\frac{dy}{dx} = \frac{x+y}{x-y}$.This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} =

Vedclass · Accepted Answer

$	an^{-1} \frac{y}{x} - \frac{1}{2} \log |x^2+y^2| = c$

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(C) Given the differential equation: $\frac{dy}{dx} = \frac{x+y}{x-y}$.This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substituting these into the equation: $v + x \frac{dv}{dx} = \frac{x+vx}{x-vx} = \frac{1+v}{1-v}$.$x \frac{dv}{dx} = \frac{1+v}{1-v} - v = \frac{1+v-v+v^2}{1-v} = \frac{1+v^2}{1-v}$.Separating the variables: $\int \frac{1-v}{1+v^2} dv = \int \frac{dx}{x}$.$\int \frac{1}{1+v^2} dv - \frac{1}{2} \int \frac{2v}{1+v^2} dv = \int \frac{dx}{x}$.Integrating both sides: $	an^{-1}(v) - \frac{1}{2} \log |1+v^2| = \log |x| + c$.Substituting $v = \frac{y}{x}$: $	an^{-1}(\frac{y}{x}) - \frac{1}{2} \log |1 + \frac{y^2}{x^2}| = \log |x| + c$.$	an^{-1}(\frac{y}{x}) - \frac{1}{2} \log |\frac{x^2+y^2}{x^2}| = \log |x| + c$.$	an^{-1}(\frac{y}{x}) - \frac{1}{2} (\log |x^2+y^2| - \log |x^2|) = \log |x| + c$.$	an^{-1}(\frac{y}{x}) - \frac{1}{2} \log |x^2+y^2| + \log |x| = \log |x| + c$.Thus,$	an^{-1}(\frac{y}{x}) - \frac{1}{2} \log |x^2+y^2| = c$.

The general solution of $\frac{dy}{dx} = \frac{x+y}{x-y}$ is

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