Question

Let $y = y(x)$ be the solution of the differential equation $\left( 2x \log_e x \right) \frac{dy}{dx} + 2y = \frac{3}{x} \log_e x, \, x>0 \, \text{and} \, y(e^{-1}) = 0.$ Then, $y(e)$ is equal to:

• A. $\frac{3}{2e}$
• B. $\frac{-2}{3e}$
• C. $\frac{-3}{e}$
• D. $\frac{-2}{e}$

Answer 1

Answer: (C)

$\frac{-3}{e}$

Answer 2

The given differential equation is:

$\frac{dy}{dx} + \frac{y}{x \ln x} = \frac{3}{2x^2}.$

Step 1: Find the integrating factor (I.F.)

$\text{I.F.} = e^{\int \frac{1}{x \ln x} dx} = e^{\ln(\ln x)} = \ln x.$

Step 2: Multiply through by I.F.

$(\ln x)y = \int \frac{3 \ln x}{2x^2} dx.$

Step 3: Solve the integral

$\int \frac{3 \ln x}{2x^2} dx = \frac{3}{2} \int x^{-2} \ln x dx.$

Use integration by parts, letting $u = \ln x$ and $dv = x^{-2} dx$:

$\int \ln x \cdot x^{-2} dx = -\frac{\ln x}{x} - \int -\frac{1}{x^2} dx = -\frac{\ln x}{x} + \frac{1}{x}.$

Thus:

$\int \frac{3 \ln x}{2x^2} dx = \frac{3}{2} \left( -\frac{\ln x}{x} + \frac{1}{x} \right).$

Step 4: Write the solution

$y \ln x = \frac{3}{2} \left( -\frac{\ln x}{x} + \frac{1}{x} \right) + C.$

Simplify:

$y = -\frac{3 \ln x}{2x \ln x} + \frac{3}{2x \ln x} + \frac{C}{\ln x}.$

$y = -\frac{3}{2x} + \frac{C}{\ln x}.$

Step 5: Apply the initial condition $y(e^{-1}) = 0$

$0 = -\frac{3}{2e^{-1}} + \frac{C}{\ln(e^{-1})}.$

$0 = -\frac{3}{2e} + \frac{C}{-1}.$

$C = -\frac{3}{2}e.$

Step 6: Find $y(e)$

$y(e) = -\frac{3}{2e} + \frac{-\frac{3}{2}e}{\ln e}.$

$y(e) = -\frac{3}{2e} - \frac{3}{2e} = -\frac{3}{e}.$

Final Answer: $-\frac{3}{e}$.