Question

In Problems 5 and 6 compute y' and y" and then combine these derivatives with y 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 $\mathit{F}\left(y\text{'},y\text{'}\text{'},y\text{'}\text{'}\text{'}\right)\mathbf{=}\mathbf{0}$. The symbols ${\mathit{c}}_{\mathbf{1}}$and ${\mathit{c}}_{\mathbf{2}}$represent constants.

$\mathit{y}\mathbf{=}{\mathit{c}}_{\mathbf{1}}{\mathit{e}}^{\mathbf{x}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}\mathit{x}{\mathit{e}}^{\mathbf{x}}$

Short answer

Answer:

The answer is $y"-2y\text{'}+y=0$.

Step-by-step solution

Define 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,

$y\text{'}=c_1{e}^x+c_{2}x{e}^2+c_2{e}^x$

Let the second derivative of the given function be,

$y"=c_1{e}^x+c_{2}x{e}^2+2c_2{e}^x$

Addon both sides of the equation and factor out 2 from the equation.

$$\begin{gathered} y"+y=2\left(c_1{e}^x+c_{2}x{e}^x\right)+2c_2{e}^x \\ =2\left(c_1{e}^x+c_{2}x{e}^x+c_2{e}^x\right) \\ =2y\text{'} \end{gathered}$$

Thus, the answer is $y"-2y\text{'}+y=0$.