Question

The general solution of the differential equation $\frac{dx}{dt} = \frac{x \log x}{t}$ is

• A. $\log x - x = c$
• B. $e^{ct} + x = 0$
• C. $\log t = x + c$
• D. $e^{ct} = x$

Hint:

Whenever an integrand contains an expression alongside its own derivative (like $\log x$ and its derivative $\frac{1}{x}$), the structure integrates into a nested logarithm: $\log(\log x)$. Exponentiating twice naturally forces a double exponential tower format, which makes option (D) the only structurally sound choice.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are required to determine the general explicit solution for the variable separation type differential equation relating $x$ and $t$.

Step 2: Key Formula or Approach:
Separate the variables to opposite sides of the equation and perform standard integrations: $$\int \frac{1}{x \log x} dx = \int \frac{1}{t} dt$$ Use substitution for the left integral: let $u = \log x \implies du = \frac{1}{x} dx$.

Step 3: Detailed Explanation:
1. Rearrange the differential equation to group variables: $$\frac{1}{x \log x} dx = \frac{1}{t} dt$$ 2. Integrate both sides: $$\int \frac{1}{x \log x} dx = \int \frac{1}{t} dt$$ 3. For the left side, substitute $u = \log x$: $$\int \frac{1}{u} du = \log|u| = \log|\log x|$$ 4. For the right side, the integration gives $\log|t|$. Combining with an arbitrary logarithmic integration constant $\log c$: $$\log(\log x) = \log t + \log c$$ $$\log(\log x) = \log(ct)$$ 5. Cancel out the outer logarithms by exponentiating both sides: $$\log x = ct$$ 6. Convert from logarithmic form to direct exponential form: $$x = e^{ct} \implies e^{ct} = x$$

Step 4: Final Answer:
The general solution of the differential equation is $e^{ct} = x$, which corresponds to option (D).