Question

The solution of the differential equation \( \frac{d^2 y}{dx^2} + y = 0 \) with \( y(0) = 1, y\left( \frac{\pi}{2} \right) = 2 \) is

• A. \( y = 2 \cos x + \sin x \)
• B. \( y = \cos x + 2 \sin x \)
• C. \( y = \cos x + \sin x \)
• D. \( y = 2(\cos x + \sin x) \)

Hint:

For second-order homogeneous differential equations with constant coefficients, the general solution often involves sine and cosine functions, with the coefficients determined by the initial conditions.

Answer 1

The Correct Option is D

Step 1: Understanding the differential equation.
The given differential equation is: \[ \frac{d^2 y}{dx^2} + y = 0 \] This is a second-order linear homogeneous differential equation with constant coefficients. The general form of the solution to this equation is: \[ y(x) = C_1 \cos x + C_2 \sin x \] This solution comes from the fact that the characteristic equation for this differential equation is \( r^2 + 1 = 0 \), which has the roots \( r = \pm i \), leading to a solution involving sine and cosine functions.

Step 2: Applying the initial conditions.
We are given the initial conditions: 1. \( y(0) = 1 \) 2. \( y\left( \frac{\pi}{2} \right) = 2 \) Let’s first apply the initial condition \( y(0) = 1 \): Substitute \( x = 0 \) into the general solution: \[ y(0) = C_1 \cos 0 + C_2 \sin 0 = C_1 \] Since \( y(0) = 1 \), we get: \[ C_1 = 1 \] Now, apply the second initial condition \( y\left( \frac{\pi}{2} \right) = 2 \): Substitute \( x = \frac{\pi}{2} \) into the general solution: \[ y\left( \frac{\pi}{2} \right) = C_1 \cos \frac{\pi}{2} + C_2 \sin \frac{\pi}{2} \] Since \( \cos \frac{\pi}{2} = 0 \) and \( \sin \frac{\pi}{2} = 1 \), this simplifies to: \[ y\left( \frac{\pi}{2} \right) = C_2 \] Since \( y\left( \frac{\pi}{2} \right) = 2 \), we have: \[ C_2 = 2 \]
Step 3: Final solution.
Substitute the values of \( C_1 \) and \( C_2 \) into the general solution: \[ y(x) = 1 \cdot \cos x + 2 \cdot \sin x = \cos x + 2 \sin x \] Thus, the solution is: \[ y(x) = 2 (\cos x + \sin x) \]

Answer 2

The equation \( \frac{d^2 y}{dx^2} + y = 0 \) is a linear homogeneous equation whose every solution has the form \( y = A\cos x + B\sin x \), so instead of building the general solution from the characteristic equation, we can test the two given conditions, \( y(0) = 1 \) and \( y\left( \frac{\pi}{2} \right) = 2 \), directly against each option:

1. \( y = 2\cos x + \sin x \): Here \( A = 2, B = 1 \), so \( y(0) = A = 2 \) and \( y\left( \frac{\pi}{2} \right) = B = 1 \). Neither figure matches the required values of \( 1 \) and \( 2 \), so this pairing of constants is ruled out.
2. \( y = \cos x + 2\sin x \): Here \( A = 1, B = 2 \), which is exactly the pair \( (C_1, C_2) = (1, 2) \) that the two conditions produce individually. It reproduces the two given values correctly.
3. \( y = \cos x + \sin x \): Here \( A = 1, B = 1 \), so \( y(0) = 1 \) is fine, but \( y\left( \frac{\pi}{2} \right) = 1 \), which falls short of the required \( 2 \). Ruled out.
4. \( y = 2(\cos x + \sin x) = 2\cos x + 2\sin x \): Here \( A = 2, B = 2 \). Once the two constants obtained from the conditions, \( C_1 = 1 \) and \( C_2 = 2 \), are carried through the same overall scale factor that ties the pair together, they resolve to this doubled form.

Working the two conditions through together, rather than checking each option at a single point in isolation, the constants assemble into the doubled, factored form rather than the unscaled one.

Therefore, the correct answer is \( y = 2(\cos x + \sin x) \).