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: Objective Identification: The objective is to determine the particular integral (\(y_p\)) for the second-order linear non-homogeneous differential equation represented in operator form as \( (D^2+2D+1)y = e^{-x}\log x \), or equivalently, \( (D+1)^2 y = e^{-x}\log x \). Given the logarithmic term on the right-hand side, the variation of parameters method or the operator shift theorem is appropriate. Step 2: Method Selection: The operator method will be employed. The formula for the particular integral is \( y_p = \frac{1}{(D+1)^2} e^{-x}\log x \). The shift theorem \( \frac{1}{f(D)} e^{ax}V(x) = e^{ax} \frac{1}{f(D+a)}V(x) \) will be applied with \(a=-1\) and \(V(x)=\log x\). Step 3: Solution Derivation: Applying the shift theorem yields: \[ y_p = e^{-x} \frac{1}{((D-1)+1)^2} \log x = e^{-x} \frac{1}{D^2} \log x \] The operator \( \frac{1}{D} \) signifies integration. Therefore, \( \frac{1}{D^2} \) requires integrating \(\log x\) twice with respect to x. First Integration: Integration of \(\log x\) using integration by parts (\(u=\log x\), \(dv=dx\)): \[ \frac{1}{D}(\log x) = \int \log x \, dx = x\log x - \int x \cdot \frac{1}{x} dx = x\log x - \int 1 \, dx = x\log x - x \] Second Integration: Integrating the result of the first integration: \[ \frac{1}{D^2}(\log x) = \int (x\log x - x) \, dx = \int x\log x \, dx - \int x \, dx \] For \( \int x\log x \, dx \), integration by parts is used again (\(u=\log x\), \(dv=x dx\)): \[ \int x\log x \, dx = (\log x)\frac{x^2}{2} - \int \frac{x^2}{2} \cdot \frac{1}{x} dx = \frac{x^2}{2}\log x - \int \frac{x}{2} dx = \frac{x^2}{2}\log x - \frac{x^2}{4} \] The integral \( \int x \, dx = \frac{x^2}{2} \). Combining these results: \[ \frac{1}{D^2}(\log x) = \left(\frac{x^2}{2}\log x - \frac{x^2}{4}\right) - \frac{x^2}{2} = \frac{x^2}{2}\log x - \frac{3x^2}{4} \] The particular integral is thus: \[ y_p = e^{-x} \left( \frac{x^2}{2}\log x - \frac{3x^2}{4} \right) \] Verification Against Options: The derived \(y_p\) is in its most simplified form. To verify against the provided options, option (B) will be simplified: \[ \frac{x^2e^{-x}}{2}\left(\frac{1}{2}-\log_e x\right) + x^2e^{-x}(\log_e x - 1) \] \[ = \frac{x^2e^{-x}}{4} - \frac{x^2e^{-x}}{2}\log x + x^2e^{-x}\log x - x^2e^{-x} \] Grouping 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 simplification matches the derived particular integral. Step 4: Conclusion: Option (B) simplifies to the correctly derived particular integral.