The solution(s) of the ordinary differential equation $y'' + y = 0$, is:
(A) $\cos x$
(B) $\sin x$
(C) $1 + \cos x$
(D) $1 + \sin x$
Choose the most appropriate answer from the options given below:
Step 1: State the differential equation.
The equation is: \[ y'' + y = 0 \]
Step 2: Determine the auxiliary equation.
The auxiliary equation is: \[ m^2 + 1 = 0 \Rightarrow m = \pm i \]
Step 3: Formulate the general solution.
The general solution to the differential equation is: \[ y(x) = C_1 \cos x + C_2 \sin x \]
Step 4: Evaluate each option.
- (A) $\cos x$: This is a solution as it aligns with the general solution.
- (B) $\sin x$: This is also a solution, fitting the general solution.
- (C) $1 + \cos x$: This is not a solution, as the constant term $1$ does not satisfy the equation.
- (D) $1 + \sin x$: This is not a solution, due to the constant term $1$ not satisfying the equation.
Step 5: Conclude.
Therefore, options (A) and (B) are the correct solutions, indicating the correct answer is (2) A and B only.
Let \( y = f(x) \) be the solution of the differential equation\[\frac{dy}{dx} + \frac{xy}{x^2 - 1} = \frac{x^6 + 4x}{\sqrt{1 - x^2}}, \quad -1 < x < 1\] such that \( f(0) = 0 \). If \[6 \int_{-1/2}^{1/2} f(x)dx = 2\pi - \alpha\] then \( \alpha^2 \) is equal to ______.
If \[ \frac{dy}{dx} + 2y \sec^2 x = 2 \sec^2 x + 3 \tan x \cdot \sec^2 x \] and
and \( f(0) = \frac{5}{4} \), then the value of \[ 12 \left( y \left( \frac{\pi}{4} \right) - \frac{1}{e^2} \right) \] equals to: