y(x)=vmatrix x x x+ x+1 27 28 27 1 1 1 vmatrix,x Rd^2ydx^2+y

Question

$y(x)=\begin{vmatrix} \sin x \quad  \cos x \quad  \sin x+\cos x+1 \ 27 \quad  28 \quad  27 \ 1 \quad  1 \quad  1 \end{vmatrix},x\in R\frac{{d}^{2}y}{d{x}^{2}}+y$

Accepted Answer

To solve for $y(x)$, we start with the determinant of the given matrix:

$\begin{vmatrix} \sin x \quad  \cos x \quad  \sin x+\cos x+1 \ 27 \quad  28 \quad  27 \ 1 \quad  1 \quad  1 \end{vmatrix}$

Using the formula for a $3	imes 3$ determinant, we expand along the first row:

$f(x)=\sin x\cdot |\begin{array}{cc} 28 & 27 \ 1 & 1 \end{array}|-\cos x\cdot |\begin{array}{cc} 27 & 27 \ 1 & 1 \end{array}|+(\sin x+\cos x+1)\cdot |\begin{array}{cc} 27 & 28 \ 1 & 1 \end{array}|$

Calculating the smaller $2	imes 2$ determinants:

$|\begin{array}{cc} 28 & 27 \ 1 & 1 \end{array}|=28\cdot 1-27\cdot 1=1$

$|\begin{array}{cc} 27 & 27 \ 1 & 1 \end{array}|=27\cdot 1-27\cdot 1=0$

$|\begin{array}{cc} 27 & 28 \ 1 & 1 \end{array}|=27\cdot 1-28\cdot 1=-1$

Substituting these into the determinant calculation gives:

$f(x)=\sin x\cdot 1-\cos x\cdot 0+(\sin x+\cos x+1)\cdot (-1)$

$f(x)=\sin x-(\sin x+\cos x+1)$

$f(x)=\sin x-\sin x-\cos x-1$

$f(x)=-\cos x-1$

Now, calculate the derivatives:

First derivative ${f}^{'}(x)$:

${f}^{'}(x)=\frac{d}{dx}(-\cos x-1)=\sin x$

Second derivative $\frac{{d}^{2}f}{d{x}^{2}}$:

$\frac{{d}^{2}f}{d{x}^{2}}=\frac{d}{dx}(\sin x)=\cos x$

Finally, compute $\frac{{d}^{2}y}{d{x}^{2}}+y$:

$\frac{{d}^{2}f}{d{x}^{2}}+f(x)=\cos x-\cos x-1=-1$

Thus, $\frac{{d}^{2}y}{d{x}^{2}}+y$ is equal to $-1$.

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