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Question

(A) Given the homogeneous differential equation $\frac{dy}{dx} = \frac{y - x}{y + x}$.Substitute $y = vx$,so $\frac{dy}{dx} = v + x\frac{dv}{dx}$.Subs

Vedclass · Accepted Answer

$\log_e(x^2 + y^2) + 2	an^{-1}\left(\frac{y}{x}ight) + c = 0$

Vedclass · Answer

(A) Given the homogeneous differential equation $\frac{dy}{dx} = \frac{y - x}{y + x}$.Substitute $y = vx$,so $\frac{dy}{dx} = v + x\frac{dv}{dx}$.Substituting into the equation: $v + x\frac{dv}{dx} = \frac{vx - x}{vx + x} = \frac{v - 1}{v + 1}$.Rearranging: $x\frac{dv}{dx} = \frac{v - 1}{v + 1} - v = \frac{v - 1 - v^2 - v}{v + 1} = -\frac{v^2 + 1}{v + 1}$.Separating variables: $\int \frac{dx}{x} = -\int \frac{v + 1}{v^2 + 1} dv$.Integrating both sides: $\log_e x = -\left[ \frac{1}{2} \int \frac{2v}{v^2 + 1} dv + \int \frac{1}{v^2 + 1} dv ight] + c$.$\log_e x = -\frac{1}{2} \log_e(v^2 + 1) - 	an^{-1}v + c$.Multiply by $-2$: $-2\log_e x = \log_e(v^2 + 1) + 2	an^{-1}v + c$.Substitute $v = \frac{y}{x}$: $-2\log_e x = \log_e\left(\frac{y^2}{x^2} + 1ight) + 2	an^{-1}\left(\frac{y}{x}ight) + c$.$-2\log_e x = \log_e\left(\frac{y^2 + x^2}{x^2}ight) + 2	an^{-1}\left(\frac{y}{x}ight) + c$.$-2\log_e x = \log_e(x^2 + y^2) - 2\log_e x + 2	an^{-1}\left(\frac{y}{x}ight) + c$.Thus,$\log_e(x^2 + y^2) + 2	an^{-1}\left(\frac{y}{x}ight) + c = 0$.

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

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