Question

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

$$\begin{gathered} \frac{dx}{dt}=2x-y \\ \frac{dy}{dt}=x \end{gathered}$$

Short answer

The solution is,

$x\left(t\right)=c_1{e}^t+c_2(t+1)e^t$

$y\left(t\right)=c_1{e}^t+c_{2}t{e}^t$

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:

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

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

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

Pull out $x$from the first equation since we have right distributive:

$$\begin{gathered} (D-2)x=-y \\ x=Dy \end{gathered}$$

The idea now is to eliminate. First operate with $D-2$ on second equation;

$$\begin{gathered} (D-2)x=-y \\ (D-2)x=\left(\right(D-2\left)D\right)z \end{gathered}$$

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

$$\begin{gathered} (D-2)D)y=-y \\ (D^2-2D)y=-y \\ (D^2-2D+1)y=0 \\ (D-1{)}^{2}y=0 \end{gathered}$$

Since the characteristic equation of this last differential equation is $(m-1{)}^2=0$, we obtain the solution:

$y\left(t\right)=c_1{e}^t+c_{2}t{e}^t$

To find the solutions.

Eliminating $y$in a similar manner yields we also get $(D-1{)}^{2}x=0$, and therefore

$x\left(t\right)=c_3{e}^t+c_{4}t{e}^t$

Now substitute $x\left(t\right)$and $y\left(t\right)$in the second equation of the first system of equations in this example,$Dy=x$:

$$\begin{gathered} D(c_1{e}^t+c_{2}t{e}^t)=c_3{e}^t+c_{4}t{e}^t \\ {c}_1{e}^t+c_2(t+1)e^t=c_3{e}^t+c_{4}t{e}^t \end{gathered}$$

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

$x\left(t\right)=c_1{e}^t+c_2(t+1)e^t$

Finally the solution is,

$x\left(t\right)=c_1{e}^t+c_2(t+1)e^t$

$y\left(t\right)=c_1{e}^t+c_{2}t{e}^t$