Question

In Problems 3 and 4 Fill in the blank and then write this result as a linear second-order differential equation that is free of the symbols ${\mathit{c}}_{\mathbf{1}}$and ${\mathit{c}}_{\mathbf{2}}$and has the form $\mathit{F}\left(y\text{'},y"\right)\mathbf{=}\mathbf{0}$. The symbols ${\mathit{c}}_{\mathbf{1}}$, ${\mathit{c}}_{\mathbf{2}}$, and k represent constants.

$\frac{{\mathbf{d}}^{\mathbf{2}}}{\mathbf{d}{\mathbf{x}}^{\mathbf{2}}}\left(c_{1}coskx+c_{2}sinkx\right)\mathbf{=}\mathit{\_}\mathit{\_}\mathit{\_}\mathit{\_}\mathit{\_}\mathit{\_}\mathit{\_}\mathit{\_}$

Short answer

Answer:

The answer is $y"+k^{2}y=0$.

Step-by-step solution

Define the second derivative of a function.

The derivative of the derivative of a function f is known as the second derivative, or second order derivative, in calculus. So, the second derivative, or the rate of change of speed with respect to time, can be used to determine the variation in speed of a car (the second derivative of distance travelled with respect to time).

Determine the second derivative of the function.

Let the first derivative of the given function be,

$$\begin{gathered} \frac{d^2}{d{x}^2}\left(c_1\cos kx+c_2\sin kx\right)=-k^2{c}_1\cos kx-k^2{c}_2\sin kx \\ =-k^2\left(c_1\cos kx+c_2\sin kx\right) \end{gathered}$$.

Let the second derivative of the given function be,

$$\begin{gathered} \frac{d^2}{d{x}^2}\left(c_1\cos kx+c_2\sin kx\right)=-k^2{c}_1\cos kx-k^2{c}_2\sin kx \\ =-k^2\left(c_1\cos kx+c_2\sin kx\right) \end{gathered}$$.

Substitute $c_1\cos kx+c_{}\sin kx=y$in the above differential equation.

$$\begin{gathered} \frac{d^{2}y}{d{x}^2}=-k^{2}y \\ \frac{d^{2}y}{d{x}^2}+k^{2}y=0 \\ y"+k^{2}y=0 \end{gathered}$$

Thus, the answer is $y"+k^{2}y=0$.