The general solution of the differential equa | vedclass

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Question

(A) Given the differential equation $\frac{d y}{d x}=\frac{3 x+y}{x-y}$.Since it is a homogeneous differential equation,let $y=vx$,then $\frac{dy}{dx}

Vedclass · Accepted Answer

$\frac{1}{\sqrt{3}} 	an ^{-1}\left(\frac{y}{x \sqrt{3}}ight)-\log \left(\frac{y^2+3 x^2}{x^2}ight)^{\frac{1}{2}}=\log (x)+C$

Vedclass · Answer

(A) Given the differential equation $\frac{d y}{d x}=\frac{3 x+y}{x-y}$.Since it is a homogeneous differential equation,let $y=vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.Substituting these into the equation:$v + x\frac{dv}{dx} = \frac{3x+vx}{x-vx} = \frac{3+v}{1-v}$.$x\frac{dv}{dx} = \frac{3+v}{1-v} - v = \frac{3+v-v+v^2}{1-v} = \frac{3+v^2}{1-v}$.Separating the variables:$\int \frac{1-v}{3+v^2} dv = \int \frac{dx}{x}$.$\int \frac{1}{3+v^2} dv - \int \frac{v}{3+v^2} dv = \int \frac{dx}{x}$.$\frac{1}{\sqrt{3}} 	an^{-1}\left(\frac{v}{\sqrt{3}}ight) - \frac{1}{2} \log(3+v^2) = \log|x| + C$.Substituting $v = \frac{y}{x}$ back:$\frac{1}{\sqrt{3}} 	an^{-1}\left(\frac{y}{x\sqrt{3}}ight) - \frac{1}{2} \log\left(3+\frac{y^2}{x^2}ight) = \log|x| + C$.$\frac{1}{\sqrt{3}} 	an^{-1}\left(\frac{y}{x\sqrt{3}}ight) - \frac{1}{2} \log\left(\frac{3x^2+y^2}{x^2}ight) = \log|x| + C$.$\frac{1}{\sqrt{3}} 	an^{-1}\left(\frac{y}{x\sqrt{3}}ight) - \log\left(\frac{y^2+3x^2}{x^2}ight)^{\frac{1}{2}} = \log|x| + C$.

The general solution of the differential equation $\frac{d y}{d x}=\frac{3 x+y}{x-y}$ is (where $C$ is a constant of integration.)

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