For the non-homogeneous second-order differential equation $\frac{d^2 y}{dx^2} + y = \sec x$, the complementary homogeneous solution is $y_c(x) = c_1 \cos x + c_2 \sin x$. Using the method of variation of parameters, a particular solution $y_p(x)$ is given by:

(A) $y_p(x) = \cos x \ln|\cos x| + x \sin x$

(B) $y_p(x) = \sin x \ln|\cos x| - x \cos x$

(C) $y_p(x) = \sec x + \tan x$

(D) $y_p(x) = x^2 \cos x$

Answer Key:

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### 1. Strategy & Theoretical Principles By the method of variation of parameters for $y'' + P(x) y' + Q(x) y = R(x)$: $$y_p(x) = u_1(x) y_1(x) + u_2(x) y_2(x)$$ where $y_1 = \cos x$, $y_2 = \sin x$, and the Wronskian is: $$W(y_1, y_2) = y_1 y_2' - y_2 y_1' = (\cos x)(\cos x) - (\sin x)(-\sin x) = \cos^2 x + \sin^2 x = 1$$ ### 2. Analytical Formulation The variable coefficients are: $$u_1(x) = -\int \frac{y_2 R(x)}{W} \, dx = -\int \frac{\sin x \sec x}{1} \, dx = -\int \tan x \, dx = \ln|\cos x|$$ $$u_2(x) = \int \frac{y_1 R(x)}{W} \, dx = \int \frac{\cos x \sec x}{1} \, dx = \int 1 \, dx = x$$ ### 3. Step-by-Step Calculation Substitute $u_1$ and $u_2$ into $y_p$: $$y_p(x) = (\ln|\cos x|) \cos x + (x) \sin x = \cos x \ln|\cos x| + x \sin x$$ ### 4. Distractor Forensics - **Option A:** Correct particular solution. - **Option B:** Swapped basis functions $y_1$ and $y_2$. - **Option C:** Guessed from trigonometric identity. ### 5. Exam Trap & Standard Reference - **Trap Alert:** Note the minus sign in $u_1 = -\int (y_2 R / W) dx$, which cancels with $-\ln|\cos x|$ from integrating $\tan x$. - **Standard Reference:** Kreyszig, E., *Advanced Engineering Mathematics*, Chapter 2.

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