Let's solve the problem **step by step** using the **method of variation of parameters**. --- ### **Given:** The inhomogeneous differential equation: \[ y'' + y = \frac{1}{\sin x} \] --- ### **Step 1: Solve the Homogeneous Equation** The homogeneous equation is: \[ y'' + y = 0 \] The general solution is: \[ y_h(x) = C_1 \cos x + C_2 \sin x \] --- ### **Step 2: Apply Variation of Parameters** We look for a particular solution of the form: \[ y_p(x) = u_1(x)\cos x + u_2(x)\sin x \] with the condition: \[ u_1'(x)\cos x + u_2'(x)\sin x = 0 \] --- ### **Step 3: Find \( u_1'(x) \) and \( u_2'(x) \)** First, compute \( y_p'(x) \): \[ y_p'(x) = u_1'(x)\cos x + u_2'(x)\sin x + u_1(x)(-\sin x) + u_2(x)\cos x \] But by the imposed condition \( u_1'(x)\cos x + u_2'(x)\sin x = 0 \): \[ y_p'(x) = -u_1(x)\sin x + u_2(x)\cos x \] Now, differentiate again for \( y_p''(x) \): \[ y_p''(x) = -u_1'(x)\sin x + u_2'(x)\cos x - u_1(x)\cos x - u_2(x)\sin x \] Now substitute into the original equation: \[ y_p''(x) + y_p(x) = \frac{1}{\sin x} \] \[ \left[ -u_1'(x)\sin x + u_2'(x)\cos x - u_1(x)\cos x - u_2(x)\sin x \right] + [u_1(x)\cos x + u_2(x)\sin x] = \frac{1}{\sin x} \] The \( u_1(x)\cos x + u_2(x)\sin x \) terms cancel: \[ -u_1'(x)\sin x + u_2'(x)\cos x = \frac{1}{\sin x} \] So, the system is: \[ \begin{cases} u_1'(x)\cos x + u_2'(x)\sin x = 0 \\ -u_1'(x)\sin x + u_2'(x)\cos x = \frac{1}{\sin x} \end{cases} \] --- ### **Step 4: Solve for \( u_2'(x) \)** Solve this linear system for \( u_2'(x) \): 1. Multiply the first equation by \(\cos x\): \[ u_1'(x)\cos^2 x + u_2'(x)\sin x \cos x = 0 \] 2. Multiply the second equation by \(\sin x\): \[ -u_1'(x)\sin^2 x + u_2'(x)\cos x \sin x = \frac{\sin x}{\sin x} = 1 \] Now subtract the first from the second: \[ (-u_1'(x)\sin^2 x + u_2'(x)\cos x \sin x) - (u_1'(x)\cos^2 x + u_2'(x)\sin x \cos x) = 1 - 0 \] \[ -u_1'(x)\sin^2 x - u_1'(x)\cos^2 x = 1 \] \[ -u_1'(x)(\sin^2 x + \cos^2 x) = 1 \] \[ -u_1'(x) = 1 \implies u_1'(x) = -1 \] From the first equation: \[ u_1'(x)\cos x + u_2'(x)\sin x = 0 \] \[ (-1)\cos x + u_2'(x)\sin x = 0 \] \[ u_2'(x) = \frac{\cos x}{\sin x} = \cot x \] --- ### **Step 5: Integrate to Find \( u_2(x) \)** \[ u_2(x) = \int \cot x \, dx = \log(\sin x) \] --- ### **Final Answer** \[ \boxed{u_2(x) = \log(\sin x)} \] This corresponds to option **(A)**.

Let's analyze each option carefully, understand what each represents, and see which one matches our derived result. --- ### **Summary of the derivation:** - The homogeneous solutions are \( y_h = C_1 \cos x + C_2 \sin x \). - Using variation of parameters, we find the particular solution involves functions \( u_1(x) \) and \( u_2(x) \). - The key step: solving for \( u_2'(x) \), which turns out to be \( \cot x \). - Integration yields: \[ u_2(x) = \int \cot x \, dx = \log |\sin x| + \text{constant} \] - For the purpose of the formula, the constant is ignored, leading to: \[ u_2(x) = \log (\sin x) \] --- ### **Now, analyze each option:** #### **(A) \( u_z(z) = \log (\sin z) \)** - **Meaning:** The derivative of \( u(z) \) with respect to \( z \) is \( \log (\sin z) \). - **Corresponds to:** Our derived \( u_2(z) \). - **Conclusion:** **Matches exactly** with our result. --- #### **(B) \( u_z(z) = -\log (\cos z) \)** - **Meaning:** The derivative of \( u(z) \) is \( -\log (\cos z) \). - **Difference:** The integral leading to this would be \( -\int \frac{1}{\cos z} dz \), which is not what we found. - **Conclusion:** **Does not match** our derivation. --- #### **(C) \( u(x) = - \frac{1}{\sin x} \)** - **Meaning:** \( u(x) \) is proportional to \( -1/\sin x \). - **Difference:** Our result is \( \log (\sin x) \), not \( -1/\sin x \). - **Conclusion:** **Incorrect**, not related to the integral we computed. --- #### **(D) \( u_z(z) = i \sin^z \)** - **Meaning:** The derivative involves a complex exponential or a different form involving \( i \). - **Difference:** No relation to our integral of \( \cot z \). - **Conclusion:** **Incorrect**. --- ### **Final Answer:** The formula corresponding to our derivation is **option (A)**: \[ \boxed{\text{Option (A): } u_z(z) = \log (\sin z)} \] This is the correct formula for the variation of parameters solution function \( u(z) \).