Question

The particular solution of the differential equation $y(1+\log x) = \left(\log x^x\right)\frac{dy}{dx}$, when $y(e) = e^2$ is

• A. $2ex \log x - y = e^2$
• B. $3ex \log y - x = 2e^2$
• C. $ex \log x + y = 2e^2$
• D. $ex \log x - y = 0$

Hint:

Recognizing functions and their derivatives instantly simplifies integration problems. Notice that the numerator $(1 + \log x)$ is the exact derivative of the denominator $(x \log x)$, meaning the integration directly evaluates to $\log|x \log x|$.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We need to solve a first-order ordinary differential equation and find its particular solution using the initial condition $y = e^2$ when $x = e$.

Step 2: Key Formula or Approach:
We first simplify the logarithmic terms using the power rule $\log x^x = x \log x$. Then, we separate the variables $x$ and $y$ onto opposite sides of the equation and integrate both sides.

Step 3: Detailed Explanation:
Given differential equation: $$y(1 + \log x) = (x \log x) \frac{dy}{dx}$$ Separating variables to group terms containing $y$ and terms containing $x$: $$\frac{1 + \log x}{x \log x} dx = \frac{1}{y} dy$$ Integrating both sides: $$\int \frac{1 + \log x}{x \log x} dx = \int \frac{1}{y} dy$$ To solve the left-hand integral, substitute $u = x \log x$. Then, using the product rule: $$du = \left(1 \cdot \log x + x \cdot \frac{1}{x}\right) dx = (\log x + 1) dx$$ The integral becomes: $$\int \frac{1}{u} du = \int \frac{1}{y} dy$$ $$\log|u| = \log|y| + \log|c|$$ $$\log|x \log x| = \log|cy| \Rightarrow x \log x = cy$$ Using the initial boundary condition $y(e) = e^2$: $$e \log(e) = c(e^2)$$ $$e(1) = c e^2 \Rightarrow c = \frac{e}{e^2} = \frac{1}{e}$$ Substitute $c = \frac{1}{e}$ back into our solution equation: $$x \log x = \frac{1}{e} y \Rightarrow ex \log x = y$$ $$ex \log x - y = 0$$

Step 4: Final Answer:
The exact particular solution matches expression (D).