The solution of frac{dy}{dx} = frac{x+y}{x-y} | vedclass

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Question

(A) 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}\left(\frac{y}{x}ight) = \log \sqrt{x^2+y^2} + C$

Vedclass · Answer

(A) 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:$\frac{1-v}{1+v^2} dv = \frac{dx}{x}$.Integrating both sides:$\int \frac{1}{1+v^2} dv - \int \frac{v}{1+v^2} dv = \int \frac{dx}{x}$.$	an^{-1}(v) - \frac{1}{2} \log(1+v^2) = \log|x| + C$.Substituting $v = \frac{y}{x}$:$	an^{-1}\left(\frac{y}{x}ight) = \log|x| + \frac{1}{2} \log\left(1 + \frac{y^2}{x^2}ight) + C$.$	an^{-1}\left(\frac{y}{x}ight) = \log|x| + \frac{1}{2} \log\left(\frac{x^2+y^2}{x^2}ight) + C$.$	an^{-1}\left(\frac{y}{x}ight) = \log|x| + \frac{1}{2} [\log(x^2+y^2) - \log(x^2)] + C$.$	an^{-1}\left(\frac{y}{x}ight) = \log|x| + \frac{1}{2} \log(x^2+y^2) - \log|x| + C$.$	an^{-1}\left(\frac{y}{x}ight) = \log\sqrt{x^2+y^2} + C$.

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

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