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: Write the given differential equation.
The equation is: \[ y'' + y = 0 \]
Step 2: Find the auxiliary equation.
The auxiliary equation is: \[ m^2 + 1 = 0 \Rightarrow m = \pm i \]
Step 3: General solution.
The general solution of the differential equation is: \[ y(x) = C_1 \cos x + C_2 \sin x \]
Step 4: Check each option.
- (A) $\cos x$: Clearly a solution, since it fits the general solution.
- (B) $\sin x$: Also a solution, since it fits the general solution.
- (C) $1 + \cos x$: Not a solution, because the constant term $1$ does not satisfy the equation.
- (D) $1 + \sin x$: Not a solution, because the constant term $1$ does not satisfy the equation.
Step 5: Conclusion.
Thus, the correct solutions are (A) and (B), making the correct answer (2) A and B only.
For the matrix, $A = \begin{bmatrix} -4 & 0 \\ -1.6 & 4 \end{bmatrix}$, the eigenvalues ($\lambda$) and eigenvectors ($X$) respectively are:
The value of $\iint_S \vec{F} \cdot \vec{N} \, ds$ where $\vec{F} = 2x^2y \hat{i} - y^2 \hat{j} + 4xz^2 \hat{k}$ and $S$ is the closed surface of the region in the first octant bounded by the cylinder $y^2 + z^2 = 9$ and the planes $x = 0, x = 2, y = 0, z = 0$, is:
The value of the integral $\displaystyle \oint_C \frac{z^3 - 6}{2z - i} \, dz$, where $C: |z| \leq 1$, is: