Question

Let \(y(x)\) be the solution of the differential equation
\( (x\log x)\frac{dy}{dx} + y = 2x\log x, \ (x \ge 1) \).
Then \(y(e)\) is equal to :

• A. 2
• B. \(2e\)
• C. \(e\)
• D. 0

Hint:

Divide by \(x\log x\) to get a linear equation; the integrating factor is \(\log x\). The domain \(x\ge 1\) forces the solution to stay finite at \(x=1\), which fixes the constant.

Answer 1

The Correct Option is A

Write \(\frac{dy}{dx} + \frac{y}{x\log x} = 2\). Integrating factor \(= e^{\int \frac{dx}{x\log x}} = e^{\log\log x} = \log x\).
So \(\frac{d}{dx}(y\log x) = 2\log x\), giving \(y\log x = 2(x\log x - x) + C\).
Since the solution is defined for \(x\ge 1\) (including \(x=1\), where \(\log x=0\)), the right side must vanish at \(x=1\): \(0 = 2(0-1)+C \Rightarrow C = 2\).
Thus \(y = 2x - \frac{2(x-1)}{\log x}\).
At \(x=e\): \(y(e) = 2e - 2(e-1) = 2\).
Answer: (1) 2