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

The Correct Option is C

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: \(-\frac{3}{e}\).

Answer 2

To solve the given differential equation, we need to find the value of \( y \) at \( x = e \). Given the differential equation:

\(\left( 2x \log_e x \right) \frac{dy}{dx} + 2y = \frac{3}{x} \log_e x\)

We can rearrange it to form a standard linear differential equation in the form of:

\(\frac{dy}{dx} + P(x)y = Q(x)\)

Where:

• \(P(x) = \frac{2}{2x \log_e x} = \frac{1}{x \log_e x}\)
• \(Q(x) = \frac{3}{2x^2 (\log_e x)^2}\)

The integrating factor (IF) is given by:

\(\text{IF} = e^{\int P(x) \, dx} = e^{\int \frac{1}{x \log_e x} \, dx}\)

Let \(u = \log_e x\), then \(\frac{du}{dx} = \frac{1}{x}\), so \(dx = x \, du\).

Thus, the integral becomes:

\(\int \frac{1}{x \log_e x} \, dx = \int \frac{1}{u} \, du = \log_e |u| + C = \log_e |\log_e x| + C\)

Therefore, the integrating factor is:

\(\text{IF} = e^{\log_e |\log_e x|} = |\log_e x|\)

We multiply the entire differential equation by the integrating factor:

\(|\log_e x|\left( \frac{dy}{dx} + \frac{y}{x \log_e x} \right) = \frac{3}{2x^2 (\log_e x)}\)

It simplifies to:

\(\frac{d}{dx}(y|\log_e x|) = \frac{3}{2x^2}\)

Integrating both sides with respect to \(x\):

\(y|\log_e x| = \int \frac{3}{2x^2} \, dx = -\frac{3}{2x} + C\)

Solving for \(y\), we get:

\(y = \frac{-3}{2x|\log_e x|} + \frac{C}{|\log_e x|}\)

Using the initial condition \(y(e^{-1}) = 0\):

\(0 = \frac{-3}{2e^{-1}|\log_e e^{-1}|} + \frac{C}{|\log_e e^{-1}|}\)

Simplifying:

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

The solving for \(C\) gives: \(C = \frac{3}{2e^{-1}}\)

Finally, finding \( y \) at \( x = e \):

\(y(e) = \frac{-3}{2e \cdot 1} + \frac{\frac{3}{2}e}{1} = \frac{-3}{2e} + \frac{3}{2e} = \frac{-3}{e}\)

Thus, the value of \( y(e) \) is \(\frac{-3}{e}\),

Therefore, the correct answer is: \(\frac{-3}{e}\)