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Question

(B) Given the differential equation: $\frac{dy}{dx} = \frac{y}{x} + \frac{\phi(y/x)}{\phi'(y/x)}$.Substitute $y = vx$,which implies $\frac{dy}{dx} = v

Vedclass · Accepted Answer

$\phi\left(\frac{y}{x}\right) = kx$

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(B) Given the differential equation: $\frac{dy}{dx} = \frac{y}{x} + \frac{\phi(y/x)}{\phi'(y/x)}$.Substitute $y = vx$,which implies $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substituting these into the equation:$v + x \frac{dv}{dx} = v + \frac{\phi(v)}{\phi'(v)}$.Subtracting $v$ from both sides:$x \frac{dv}{dx} = \frac{\phi(v)}{\phi'(v)}$.Separating the variables:$\frac{\phi'(v)}{\phi(v)} dv = \frac{dx}{x}$.Integrating both sides:$\int \frac{\phi'(v)}{\phi(v)} dv = \int \frac{dx}{x}$.$\ln|\phi(v)| = \ln|x| + C$,where $C = \ln|k|$.$\ln|\phi(v)| = \ln|kx|$.Taking the exponential of both sides:$\phi(v) = kx$.Substituting $v = \frac{y}{x}$ back:$\phi\left(\frac{y}{x}\right) = kx$.

The solution of the differential equation $\frac{dy}{dx} = \frac{y}{x} + \frac{\phi(y/x)}{\phi'(y/x)}$ is

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