Question

If the solution curve of the differential equation \( (y - 2 \log x) dx + (x \log x^2) dy = 0 \) passes through the points \( \left( e^{4/3}, \alpha \right) \) and \( \left( e^4, \alpha \right) \), then \( \alpha \) is equal to: 
 

Hint:

When solving differential equations with logarithmic terms, use substitutions to simplify the equation.

Answer 1

Correct Answer: 3

Given the differential equation: \[ (y - 2 \log x) dx + (x \log x^2) dy = 0, \] we first rearrange it as: \[ dy = \frac{1}{x \log x^2} \cdot \left[ (2 \log x - y) \right] dx. \] Next, calculate the integrating factor (I.F.): \[ I.F. = e^{\int \frac{1}{x \log x^2} dx}. \] We simplify this integral: \[ I.F. = e^{\int \frac{1}{2x \ln x} dx}. \] Let \( \ln x = u \), so that \( \frac{1}{x} dx = du \): \[ I.F. = e^{\int \frac{1}{2 \ln x} dx} = e^{\frac{1}{2} \ln (\ln x)}. \] Thus, the I.F. is \( (\ln x)^{1/2} \). Now, multiply the original equation by the I.F.: \[ y \cdot \ln x = \int \sqrt{\ln x} \, dx. \] Simplifying further: \[ \int \sqrt{\ln x} \, dx = 2 \int u^2 du = \frac{2}{3} u^3. \] Substitute back \( u = \ln x \): \[ y \cdot \ln x = \frac{2}{3} (\ln x)^3 + C. \] Substitute the known points \( \left( e^{4/3}, \frac{4}{3} \right) \) and \( \left( e^4, \alpha \right) \) into the equation: \[ \frac{4}{3} \cdot \ln e^{4/3} = \frac{2}{3} (\ln e^{4/3})^3 + C, \] which simplifies to: \[ \frac{4}{3} \cdot \frac{4}{3} = \frac{2}{3} \cdot \left( \frac{4}{3} \right)^3 + C \quad \Rightarrow \quad C = \frac{2}{3}. \] Now, substitute the second point \( (e^4, \alpha) \): \[ \alpha \cdot \ln e^4 = \frac{2}{3} \left( \ln e^4 \right)^3 + \frac{2}{3}, \] which simplifies to: \[ \alpha \cdot 4 = \frac{2}{3} \cdot 64 + \frac{2}{3} = \frac{128}{3} + \frac{2}{3} = \frac{130}{3}. \] Thus: \[ \alpha = \frac{130}{3} \times \frac{1}{4} = \frac{65}{6} \quad \Rightarrow \quad \boxed{3}. \]