Question

In Problems 41 and 42 verify that the indicated pair of functions is a solution of the given system of differential equations on the interval .

$$\begin{gathered} \frac{d^{2}x}{d{t}^2}=4y+e^t\backslash \\ \frac{d^{2}y}{d{t}^2}=4x-e^t; \\ x=\cos 2t+\sin 2t+\frac{1}{5}{e}^t, \\ y=-\cos 2t-\sin 2t-\frac{1}{5}{e}^t \end{gathered}$$

Short answer

For all the entire interval $(-\infty ,\infty )$ which is true, so we have verified that the pair of equations, $x=\cos 2t+\sin 2t+\frac{1}{5}{e}^t$and $y=-\cos 2t-\sin 2t-\frac{1}{5}{e}^t$is a solution to the system of differential equations.

Step-by-step solution

Definition of differential equation.

A differential equation is defined as the derivative function of one or more unknown functions (dependable variables) with respect to one or more undependable variables.

First and second derivative of x .

Given$x=\cos 2t+\sin 2t+\frac{1}{5}{e}^t$

Finding the first derivation of,

$x\text{'}=-2\sin 2t+2\cos 2t+\frac{1}{5}{e}^t$

Finding the second derivation of,

$X"=-4\cos 2t-4\sin 2t+\frac{1}{5}{e}^t$

First and second derivative of y .

Given that,$y=-\cos 2t-\sin 2t-\frac{1}{5}{e}^t$

Finding the first derivation of$y$,

$y\text{'}=2\sin 2t-2\cos 2t-\frac{1}{5}{e}^t$

Finding the second derivation of$y$,

$y"=4\cos 2t+4\sin 2t-\frac{1}{5}{e}^t$

Substitute the values into the system of equations.

$-4\cos 2t-4\sin 2t+\frac{1}{5}{e}^t=4y+e^t$.....(1)

$2\sin 2t-2\cos 2t-\frac{1}{5}{e}^t=4x-e^t$....(2)

Add the equations (1) and (2), we get

$0+0+0=4y+4x+0$

Remember that$y=-x$, so substitute that in above equation,

role="math" localid="1664210203399" $$\begin{gathered} 4y+4(-y)=0 \\ 4y=4y \end{gathered}$$

Hence, For all the entire interval $(-\infty ,\infty )$ which is true, so we have verified that the pair of equations, $x=\cos 2t+\sin 2t+\frac{1}{5}{e}^t$ and $y=-\cos 2t-\sin 2t-\frac{1}{5}{e}^t$is a solution to the system of differential equations.