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EEJ MAIN Mathematics QUESTION #6004
Question 1

Let $y(x)$ be the solution of the differential equation $(x \log x)\dfrac{dy}{dx} + y = 2x \log x$, where $x \geq 1$. Find the value of $y(e)$.

  • $e$
  • 2✔️
  • $2e$
Correct Answer Explanation

Rewrite in standard linear form: $\frac{dy}{dx} + \frac{y}{x\log x} = 2$

Integrating factor: $e^{\int \frac{1}{x\log x}dx} = e^{\ln(\log x)} = \log x$

Multiply through: $\frac{d}{dx}(y \log x) = 2\log x$

Integrate: $y\log x = 2(x\log x - x) + C$

At $x=1$: $y(1)\cdot 0 = 2(0-1)+C \Rightarrow C=2$

So $y\log x = 2x\log x - 2x + 2$. At $x=e$: $y(1) = 2e - 2e + 2 = \mathbf{2}$