Question

The general solution to the ordinary differential equation \( \dfrac{d^2y}{dx^2} = \cos 2x \) is

• A. \( y = -\dfrac{\cos 2x}{4} + C_1 x + C_2 \), where \( C_1 \) and \( C_2 \) are constants
• B. \( y = \dfrac{\sin 2x}{2} + C_1 \), where \( C_1 \) is a constant
• C. \( y = \dfrac{\cos 2x}{4} + C_1 x + C_2 \), where \( C_1 \) and \( C_2 \) are constants
• D. \( y = -\dfrac{\sin 2x}{2} + C_1 \), where \( C_1 \) is a constant

Hint:

Integrate both sides of the equation twice with respect to x, adding a constant each time.

Answer 1

The Correct Option is A

Step 1: Integrate once.
Start from \( \dfrac{d^2y}{dx^2} = \cos 2x \).
Integrate both sides with respect to \( x \): \( \dfrac{dy}{dx} = \displaystyle\int \cos 2x \, dx = \dfrac{\sin 2x}{2} + C_1 \).

Step 2: Integrate a second time.
\( y = \displaystyle\int \left( \dfrac{\sin 2x}{2} + C_1 \right) dx \)
\( y = -\dfrac{\cos 2x}{4} + C_1 x + C_2 \), where \( C_2 \) is the new constant from this integration.

Step 3: Check by differentiating back.
\( \dfrac{dy}{dx} = \dfrac{\sin 2x}{2} + C_1 \), and \( \dfrac{d^2y}{dx^2} = \cos 2x \), which matches the original equation.

Final Answer:
The general solution is \( y = -\dfrac{\cos 2x}{4} + C_1 x + C_2 \). \[ \boxed{y = -\tfrac{\cos 2x}{4} + C_1 x + C_2} \]