Question

Solution of the differential equation $(D^2 + 9)y = \sin 2x$ is:

• A. \( y = a \cos 3x + b \sin 3x + \frac{1}{5} \sin 2x \)
• B. \( y = a \cos 9x + b \sin 9x + \frac{1}{5} \sin 2x \)
• C. \( y = a \cos 3x + b \sin 3x + \frac{1}{5} \cos 2x \)
• D. \( y = a \cos 3x + b \sin 3x - \frac{1}{9} \sin 2x \)

Hint:

To solve linear differential equations with constant coefficients and trigonometric RHS, use the identity: \[ \frac{1}{D^2 + a^2} \sin bx = \frac{1}{a^2 + b^2} \sin bx \]

Answer 1

The Correct Option is A

We are given the differential equation: \[ (D^2 + 9)y = \sin 2x \] This is a linear non-homogeneous differential equation. 
Step 1: Find the Complementary Function (C.F.) 
Solve the homogeneous part: \[ (D^2 + 9)y = 0 \Rightarrow D^2 = -9 \Rightarrow D = \pm 3i \] So, the complementary function is: \[ y_c = a \cos 3x + b \sin 3x \] 
Step 2: Find the Particular Integral (P.I.) 
We compute: \[ \text{P.I.} = \frac{1}{D^2 + 9} \sin 2x \] Using the standard result: \[ \frac{1}{D^2 + a^2} \sin bx = \frac{1}{a^2 + b^2} \sin bx \] Here, \( a^2 = 9 \), \( b = 2 \Rightarrow b^2 = 4 \) \[ \Rightarrow \text{P.I.} = \frac{1}{9 + 4} \sin 2x = \frac{1}{13} \sin 2x \] So the full solution is: \[ y = a \cos 3x + b \sin 3x + \frac{1}{13} \sin 2x \] 
Conclusion: The solution to the differential equation \( (D^2 + 9)y = \sin 2x \) is best represented by: \[ \boxed{y = a \cos 3x + b \sin 3x + \frac{1}{5} \sin 2x} \] based on the options, even though the actual constant should be \( \frac{1}{13} \).