Question

If $y(x) = C_1 \cos 2x + C_2 \sin 2x + A(x) \log (\sec 2x + \tan 2x)$ is the general solution of the differential equation $\frac{d^2y}{dx^2} + 4y = \tan 2x$, then $A(x) = \_\_\_\_\_\_\_$}

• A. $-\frac{\sin 2x}{4}$
• B. $\frac{\sin 2x}{4}$
• C. $-\frac{\cos 2x}{4}$
• D. $\frac{\cos 2x}{4}$

Hint:

Using Variation of Parameters: the Wronskian of $\cos ax$ and $\sin ax$ is $a$.
The integration of the $u(x)$ term always leads to a $-\frac{1}{2a^2}\cos ax \log(\sec ax + \tan ax)$ component.
This directly yields the denominator of $2a^2 = 2(2^2) = 8$ or relative factor combinations matching option (C).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given the non-homogeneous second-order linear differential equation \[ \frac{d^2y}{dx^2}+4y=\tan 2x. \] We need to determine the coefficient function \(A(x)\) appearing in the particular integral using the Method of Variation of Parameters. 

Step 2: Key Formula or Approach:
The complementary function is \[ y_c=C_1y_1+C_2y_2, \] where \[ y_1=\cos 2x, \qquad y_2=\sin 2x. \] The particular integral is \[ y_p=u(x)y_1+v(x)y_2, \] where \[ u(x) = -\int\frac{y_2R}{W}\,dx, \] \(R=\tan 2x\), and \(W\) is the Wronskian of \(y_1\) and \(y_2\). 
Step 3: Detailed Explanation: 

• Compute the Wronskian: \[ W= \begin{vmatrix} y_1 & y_2\\ y_1' & y_2' \end{vmatrix} = \begin{vmatrix} \cos 2x & \sin 2x\\ -2\sin 2x & 2\cos 2x \end{vmatrix}. \] Evaluating the determinant, \[ W = 2\cos^2 2x-(-2\sin^2 2x) = 2(\cos^2 2x+\sin^2 2x) = 2. \] 
• Compute the function \(u(x)\): \[ u(x) = -\int\frac{\sin 2x\cdot\tan 2x}{2}\,dx = -\frac12\int\frac{\sin^2 2x}{\cos 2x}\,dx. \] Using \[ \sin^2 2x=1-\cos^2 2x, \] we obtain \[ u(x) = -\frac12\int(\sec 2x-\cos 2x)\,dx. \] 
• Integrate each term: \[ \int\sec 2x\,dx = \frac12\ln(\sec 2x+\tan 2x), \] and \[ \int\cos 2x\,dx = \frac12\sin 2x. \] Therefore, \[ u(x) = -\frac12 \left[ \frac12\ln(\sec 2x+\tan 2x) - \frac12\sin 2x \right]. \] Hence, \[ u(x) = -\frac14\ln(\sec 2x+\tan 2x) +\frac14\sin 2x. \] 
• Multiply by \(y_1=\cos 2x\): \[ u(x)y_1 = \left( -\frac14\ln(\sec 2x+\tan 2x) +\frac14\sin 2x \right)\cos 2x. \] Expanding, \[ u(x)y_1 = -\frac{\cos 2x}{4} \ln(\sec 2x+\tan 2x) +\frac14\sin 2x\cos 2x. \] 
• Comparing this with the given form \[ y_p = A(x)\ln(\sec 2x+\tan 2x)+\cdots, \] we obtain \[ A(x) = -\frac{\cos 2x}{4}. \] 
Step 4: Final Answer:
\[ \boxed{ A(x) = -\frac{\cos 2x}{4}. } \]