The differential equation y^{prime} = frac{y} | vedclass

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(C) Given the differential equation $\frac{dy}{dx} = \frac{y}{x + \sqrt{xy}}$.Since it is a homogeneous differential equation,let $y = vx$,then $\frac

Vedclass · Accepted Answer

$2\sqrt{x/y} = \ln|x| + C$

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(C) Given the differential equation $\frac{dy}{dx} = \frac{y}{x + \sqrt{xy}}$.Since it is a homogeneous differential equation,let $y = vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.Substituting these into the equation: $v + x\frac{dv}{dx} = \frac{vx}{x + \sqrt{x^2v}} = \frac{vx}{x(1 + \sqrt{v})} = \frac{v}{1 + \sqrt{v}}$.$x\frac{dv}{dx} = \frac{v}{1 + \sqrt{v}} - v = \frac{v - v - v\sqrt{v}}{1 + \sqrt{v}} = \frac{-v\sqrt{v}}{1 + \sqrt{v}}$.Separating variables: $-\frac{1 + \sqrt{v}}{v\sqrt{v}} dv = \frac{dx}{x} \Rightarrow -(\frac{1}{v^{3/2}} + \frac{1}{v}) dv = \frac{dx}{x}$.Integrating both sides: $-\int (v^{-3/2} + v^{-1}) dv = \int \frac{1}{x} dx$.$-(-2v^{-1/2} - \ln|v|) = \ln|x| + C$.$2\frac{1}{\sqrt{v}} + \ln|v| = \ln|x| + C$.Since $v = y/x$,we have $2\sqrt{x/y} + \ln(y/x) = \ln|x| + C$.$2\sqrt{x/y} + \ln|y| - \ln|x| = \ln|x| + C$.$2\sqrt{x/y} = \ln|x| - \ln|y| + \ln|x| + C = 2\ln|x| - \ln|y| + C$.

The differential equation $y^{\prime} = \frac{y}{x + \sqrt{xy}}$ has a general solution given by (where $C$ is a constant of integration):

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