Question

If a curve \(y = f(x)\), passing through the point \((1, 2)\), is the solution of the differential equation, \(2x^2\,dy = (2xy + y^2)\,dx\), then \(f\left(\frac{1}{2}\right)\) is equal to :

• A. \(\frac{1}{1-\log_e 2}\)
• B. \(\frac{1}{1+\log_e 2}\)
• C. \(\frac{-1}{1+\log_e 2}\)
• D. \(1+\log_e 2\)

Hint:

The equation is homogeneous: put \(y=vx\), separate the variables and use \(y(1)=2\).

Answer 1

The Correct Option is B

\(\frac{dy}{dx}=\frac{2xy+y^2}{2x^2}\). Put \(y=vx\), \(\frac{dy}{dx}=v+x\frac{dv}{dx}\).
\(v+x\frac{dv}{dx} = v+\frac{v^2}{2}\Rightarrow x\frac{dv}{dx}=\frac{v^2}{2}\Rightarrow \frac{2\,dv}{v^2}=\frac{dx}{x}\).
Integrating: \(-\frac{2}{v} = \ln x + C\).
At \((1,2)\): \(v=2\), so \(-1 = 0 + C\Rightarrow C=-1\). Thus \(-\frac{2x}{y} = \ln x - 1\Rightarrow y=\frac{2x}{1-\ln x}\).
At \(x=\frac12\): \(y = \frac{1}{1-\ln\frac12} = \frac{1}{1+\ln 2}\).
Answer: (2) \(\frac{1}{1+\log_e 2}\)