Question

If y=\(\frac{x}{log_e|cx|}\) is the solution of the differential equation \(\frac{dy}{dx}=\frac{y}{x}+\phi(\frac{x}{y})\), then \(\phi (\frac{x}{y})\) is given by 

• A. \(\frac{y^2}{x^2}\)
• B. \(-\frac{y^2}{x^2}\)
• C. \(\frac{x^2}{y^2}\)
• D. \(-\frac{x^2}{y^2}\)

Answer 1

The Correct Option is A

We are given the differential equation: \[ \frac{dy}{dx} = xy + \phi\left(\frac{y}{x}\right) \] Let’s use the substitution \( y = vx \), so that: \[ \frac{dy}{dx} = v + x\frac{dv}{dx} \] Substitute into the original equation: \[ v + x\frac{dv}{dx} = x(v) + \phi(v) \Rightarrow v + x\frac{dv}{dx} = xv + \phi(v) \] Cancel \(v\) from both sides: \[ x\frac{dv}{dx} = \phi(v) \Rightarrow \frac{dv}{\phi(v)} = \frac{dx}{x} \] Integrating both sides: \[ \int \frac{1}{\phi(v)} dv = \int \frac{1}{x} dx = \ln x + C \] Let the left side be some function \(F(v)\), so: \[ F(v) = \ln x + C \Rightarrow x = k e^{F(v)} \] Now since \(v = \frac{y}{x}\), then \(F\left(\frac{y}{x}\right) = \ln x + C\) From this, the general solution will involve the expression: \[ F\left(\frac{y}{x}\right) - \ln x = \text{constant} \] Which means the solution depends on: \[ \frac{y}{x} \quad \text{or} \quad \frac{y^2}{x^2} \] Therefore, the correct answer is: Option (A): \( \frac{y^2}{x^2} \)