Question

The general solution of the differential equation $y(1+\log x)\left(\frac{dx}{dy}\right) - x\log x = 0$ is

• A. $y(1+\log x) = c$
• B. $x\log x = yc$
• C. $x\log x = y+c$
• D. $\log x - y = c$

Hint:

To easily spot logarithmic substitutions: if you see an expression and its exact derivative appearing together in a fraction, the integral of $\frac{f'(x)}{f(x)} dx$ is always simply $\log|f(x)|$. No complex u-substitution required if you recognize this pattern!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are asked to solve a first-order differential equation. By looking at the terms, we can see that all $x$ terms can be grouped with $dx$ and all $y$ terms with $dy$, meaning it is a variable-separable differential equation.

Step 2: Key Formula or Approach:
Rearrange the equation to isolate $x$ terms on one side and $y$ terms on the other: $f(x)dx = g(y)dy$.
Then, integrate both sides.
For the $x$-integration, use the substitution method: let $u = x\log x$.

Step 3: Detailed Explanation:
Start with the given equation:
$$y(1+\log x)\frac{dx}{dy} - x\log x = 0$$ Move the second term to the right side:
$$y(1+\log x)\frac{dx}{dy} = x\log x$$ Separate the variables by dividing both sides by $(x\log x \cdot y)$ and multiplying by $dy$:
$$\frac{1+\log x}{x\log x} dx = \frac{1}{y} dy$$ Integrate both sides:
$$\int \frac{1+\log x}{x\log x} \, dx = \int \frac{1}{y} \, dy$$ For the left-hand side integral, let $u = x\log x$.
Differentiate using the product rule: $du = \left( 1 \cdot \log x + x \cdot \frac{1}{x} \right) dx = (\log x + 1) dx$.
Notice that the numerator $(\log x + 1)dx$ perfectly matches $du$.
Substitute into the integral:
$$\int \frac{1}{u} \, du = \int \frac{1}{y} \, dy$$ $$\log|u| = \log|y| + \log|c|$$ Using the property $\log m + \log n = \log(mn)$:
$$\log|u| = \log|yc|$$ Drop the logarithms:
$$u = yc$$ Substitute $u$ back with $x\log x$:
$$x\log x = yc$$

Step 4: Final Answer:
The general solution is $x\log x = yc$, matching option (B).