If $y(x) = \begin{vmatrix} \sin x & \cos x & \sin x + \cos x + 1 \\ 27 & 28 & 27 \\ 1 & 1 & 1 \end{vmatrix}$, $x \in R$, then $\frac{d^2y}{dx^2} + y$ is equal to
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Determinant Function Derivative Calculation
The given function is defined by a determinant:
$ y(x) = \begin{vmatrix} \sin x & \cos x & \sin x + \cos x + 1 \\ 27 & 28 & 27 \\ 1 & 1 & 1 \end{vmatrix} $
Simplify Determinant
To simplify the determinant, apply the row operation $R_2 \to R_2 - 27 R_3$. This operation does not change the value of the determinant.
The second row \( [27, 28, 27] \) transforms as follows:
- \( [27, 28, 27] - 27 \times [1, 1, 1] = [27-27, 28-27, 27-27] = [0, 1, 0] \)
The determinant simplifies to:
$ y(x) = \begin{vmatrix} \sin x & \cos x & \sin x + \cos x + 1 \\ 0 & 1 & 0 \\ 1 & 1 & 1 \end{vmatrix} $
Expand this determinant along the second row ($R_2$) for easier calculation:
$ y(x) = -1 \cdot \begin{vmatrix} \sin x & \sin x + \cos x + 1 \\ 1 & 1 \end{vmatrix} $
Calculate the remaining 2x2 determinant:
$ y(x) = -1 \left( (\sin x)(1) - (\sin x + \cos x + 1)(1) \right) $
$ y(x) = -1 \left( \sin x - \sin x - \cos x - 1 \right) $
$ y(x) = -1 \left( -\cos x - 1 \right) $
$ y(x) = \cos x + 1 $
Calculate Derivatives
Now, find the first and second derivatives of \( y(x) = \cos x + 1 \).
The first derivative is:
$ \frac{dy}{dx} = \frac{d}{dx}(\cos x + 1) = -\sin x $
The second derivative is:
$ \frac{d^2y}{dx^2} = \frac{d}{dx}(-\sin x) = -\cos x $
Final Expression Evaluation
Finally, evaluate the expression \( \frac{d^2y}{dx^2} + y \):
$ \frac{d^2y}{dx^2} + y = (-\cos x) + (\cos x + 1) $
Combine like terms:
$ \frac{d^2y}{dx^2} + y = -\cos x + \cos x + 1 $
$ \frac{d^2y}{dx^2} + y = 1 $
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