Question

In Problems 1–20 solve the given system of differential equations bysystematic elimination.

$\begin{array}{rcl}\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}& \mathbf{=}& \mathbf{4}\mathit{x}\mathbf{+}\mathbf{7}\mathit{y}\\ \frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{t}}& \mathbf{=}& \mathit{x}\mathbf{-}\mathbf{2}\mathit{y}\end{array}$

Short answer

The solution is,

$x\left(t\right)=7c_1{e}^{5t}-c_2{e}^{-3t}$

$y\left(t\right)=c_1{e}^{5t}+c_2{e}^{-3t}$

Step-by-step solution

Step 1:Definition of differential equation.

An equation containing the derivatives of one or more unknown functions (or dependent variables), with respect to one or more independent variables, is said to be a differential equation (DE).

Differential operator notation.

First we write the system in differential operator notation:

$$\begin{gathered} Dx=4x+7y \\ Dy=x-2y \end{gathered}$$

Switch derivatives of $x$on left side and derivatives of $y$on right, in both of these equations:

$$\begin{gathered} Dx-4x=7y \\ x=Dy+2y \end{gathered}$$

Pull out $x$from the first equation and y from the second since we have right distributivity:

$$\begin{gathered} (D-4)x=7y \\ x=(D+2)y \end{gathered}$$

The idea now is to eliminate. First operate with $D-4$on second equation;

$$\begin{gathered} (D-4)x=7y \\ (D-4)x=(D-4)(D+2)y \end{gathered}$$

Now since left sides of each equation is equal, also right sides are equal:

$$\begin{gathered} (D-4)(D+2)y=7y \\ (D^2+2D-4D-8)y=7y \\ (D^2-2D-8-7)y=0 \\ (D-5)(D+3)y=0 \end{gathered}$$

Since the characteristic equation of this last differential equation is $(m-5)(m+3)=0$, we obtain the solution:

$y\left(t\right)=c_1{e}^{5t}+c_2{e}^{-3t}$

To find the solutions:

Eliminating y in a similar manner yields we also get$(D-5)(D+3)x=0$, and therefore

$x\left(t\right)=c_3{e}^{5t}+c_4{e}^{-3t}$

Now substitute $x\left(t\right)$and $y\left(t\right)$ in the equation from above,

$$\begin{gathered} (D+2)(c_1{e}^{5t}+c_2{e}^{-3t})=c_3{e}^{5t}+c_4{e}^{-3t} \\ 5c_1{e}^{5t}-3c_2{e}^{-3t}+2(c_1{e}^{5t}+c_2{e}^{-3t})=c_3{e}^{5t}+c_4{e}^{-3t} \\ 7c_1{e}^{5t}-c_2{e}^{-3t}=c_3{e}^{5t}+c_4{e}^{-3t} \end{gathered}$$

Therefore, we can express $x\left(t\right)$in terms of $c_1$and $c_2$:

$x\left(t\right)=7c_1{e}^{5t}-c_2{e}^{-3t}$

Finally the solution is,

$x\left(t\right)=7c_1{e}^{5t}-c_2{e}^{-3t}$

$y\left(t\right)=c_1{e}^{5t}+c_2{e}^{-3t}$