The general solution of the differential equa | vedclass

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(B) The given differential equation is $\frac{dy}{dx} = \frac{x+y}{x-y}$. This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx

Vedclass · Accepted Answer

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

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(B) The given differential equation is $\frac{dy}{dx} = \frac{x+y}{x-y}$. This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$. Substituting this 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 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) - \frac{1}{2} \log\left(1+\frac{y^2}{x^2}ight) = \log|x| + C$. $	an^{-1}\left(\frac{y}{x}ight) = \frac{1}{2} \log\left(\frac{x^2+y^2}{x^2}ight) + \log|x| + C = \frac{1}{2} \log(x^2+y^2) - \log|x| + \log|x| + C$. $	an^{-1}\left(\frac{y}{x}ight) = \log\left(\sqrt{x^2+y^2}ight) + C$. Thus,$	an^{-1}\left(\frac{y}{x}ight) = \log\left(c\sqrt{x^2+y^2}ight)$.

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

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