Question

If \(y(x)\) is the solution of the differential equation \(\frac{dy}{dx} + \left(\frac{2x+1}{x}\right)y = e^{-2x},\ x \gt 0\), where \(y(1) = \frac{1}{2}e^{-2}\), then

• A. \(y(\log_e 2) = \log_e 4\)
• B. \(y(x)\) is decreasing in \((0, 1)\)
• C. \(y(\log_e 2) = \frac{\log_e 2}{4}\)
• D. \(y(x)\) is decreasing in \(\left(\frac{1}{2}, 1\right)\)

Hint:

Use the integrating factor \(x e^{2x}\); then study the sign of \(y'\) and evaluate \(y(\ln 2)\).

Answer 1

The Correct Option is D

Integrating factor: \(e^{\int (2 + 1/x)dx} = e^{2x + \ln x} = x e^{2x}\).
\(\frac{d}{dx}\left(y\,x e^{2x}\right) = x e^{2x}\cdot e^{-2x} = x \Rightarrow y\,x e^{2x} = \frac{x^2}{2} + C\).
\(y(1) = \frac12 e^{-2}\): \(\frac12 e^{-2}\cdot e^{2} = \frac12 + C \Rightarrow C = 0\). So \(y = \frac{x}{2}e^{-2x}\).
\(y(\ln2) = \frac{\ln 2}{2}\cdot\frac14 = \frac{\ln2}{8}\), so options (1) and (3) are false.
\(y' = \frac12 e^{-2x}(1 - 2x)\): positive for \(x \lt \frac12\) and negative for \(x \gt \frac12\). So y increases on \((0,\frac12)\) and decreases on \((\frac12, 1)\); option (2) is false and option (4) is true.
Answer: (4) \(y(x)\) is decreasing in \(\left(\frac{1}{2}, 1\right)\)