Question

The particular integral of differential equation \( \frac{d^2y}{dx^2} + 2\frac{dy}{dx} + y = e^{-x}\log x \) is:

• A. \( \frac{x^2}{2}\left(\frac{1}{2}-\log_e x\right)e^{-x} + e^{-2x}(x\log_e x - x) \)
• B. \( \frac{x^2e^{-x}}{2}\left(\frac{1}{2}-\log_e x\right) + x^2e^{-x}(\log_e x - 1) \)
• C. \( \frac{x^2}{2}\left(\frac{1}{3}-\log_e x\right)e^{-x} + e^{-x}(x\log_e x - x) \)
• D. \( \frac{x^2}{2}\left(\frac{1}{3}-\log_e x\right)e^{-x} + x^2e^{-x}(\log_e x - 1) \)

Hint:

The operator shift theorem is very powerful for particular integrals where the forcing function is of the form \(e^{ax}V(x)\). It reduces the problem to finding the particular integral for the function \(V(x)\) with a simpler operator. For repeated roots in the complementary function, like \((D+a)^2 y = ...\), the method of variation of parameters is also a reliable, albeit sometimes longer, alternative.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We need to find the particular integral (\(y_p\)) for a second-order linear non-homogeneous differential equation. The equation can be written in operator form as \( (D^2+2D+1)y = e^{-x}\log x \), or \( (D+1)^2 y = e^{-x}\log x \). Since the right-hand side involves a logarithmic term, the method of variation of parameters or the operator shift theorem is suitable.
Step 2: Key Formula or Approach:
We will use the operator method. The particular integral is given by \( y_p = \frac{1}{(D+1)^2} e^{-x}\log x \). We use the shift theorem: \( \frac{1}{f(D)} e^{ax}V(x) = e^{ax} \frac{1}{f(D+a)}V(x) \). Here, \(a=-1\) and \(V(x)=\log x\).
Step 3: Detailed Explanation:
Applying the shift theorem: \[ y_p = e^{-x} \frac{1}{((D-1)+1)^2} \log x = e^{-x} \frac{1}{D^2} \log x \] The operator \( \frac{1}{D} \) represents integration. Therefore, \( \frac{1}{D^2} \) means we need to integrate \(\log x\) twice with respect to x.
First Integration: We use integration by parts for \( \int \log x \, dx \): Let \(u=\log x\) and \(dv=dx\). Then \(du = \frac{1}{x}dx\) and \(v=x\). \[ \frac{1}{D}(\log x) = \int \log x \, dx = x\log x - \int x \cdot \frac{1}{x} dx = x\log x - \int 1 \, dx = x\log x - x \] Second Integration: \[ \frac{1}{D^2}(\log x) = \int (x\log x - x) \, dx = \int x\log x \, dx - \int x \, dx \] For \( \int x\log x \, dx \), we use integration by parts again: Let \(u=\log x\) and \(dv=x dx\). Then \(du = \frac{1}{x}dx\) and \(v=\frac{x^2}{2}\). \[ \int x\log x \, dx = (\log x)\frac{x^2}{2} - \int \frac{x^2}{2} \cdot \frac{1}{x} dx = \frac{x^2}{2}\log x - \int \frac{x}{2} dx = \frac{x^2}{2}\log x - \frac{x^2}{4} \] The second term is \( \int x \, dx = \frac{x^2}{2} \). Combining these: \[ \frac{1}{D^2}(\log x) = \left(\frac{x^2}{2}\log x - \frac{x^2}{4}\right) - \frac{x^2}{2} = \frac{x^2}{2}\log x - \frac{3x^2}{4} \] Now, we find \(y_p\): \[ y_p = e^{-x} \left( \frac{x^2}{2}\log x - \frac{3x^2}{4} \right) \] Comparing with the options: The derived solution is in its simplest form. The options are not. We must simplify the correct option to see if it matches our result. Let's simplify option (B): \[ \frac{x^2e^{-x}}{2}\left(\frac{1}{2}-\log_e x\right) + x^2e^{-x}(\log_e x - 1) \] \[ = \left(\frac{x^2e^{-x}}{4} - \frac{x^2e^{-x}}{2}\log x\right) + (x^2e^{-x}\log x - x^2e^{-x}) \] Group the terms: \[ = e^{-x}\left(\frac{x^2}{4} - x^2\right) + e^{-x}\log x\left(-\frac{x^2}{2} + x^2\right) \] \[ = e^{-x}\left(-\frac{3x^2}{4}\right) + e^{-x}\log x\left(\frac{x^2}{2}\right) \] \[ = e^{-x}\left(\frac{x^2}{2}\log x - \frac{3x^2}{4}\right) \] This matches our derived particular integral. Step 4: Final Answer:
The particular integral is given by option (B), which simplifies to the correctly derived expression.