Question

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

$$\begin{gathered} \mathbf{13}\mathbf{.}\mathbf{}\mathbf{2}\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{-}\mathbf{5}\mathit{x}\mathbf{+}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{t}}\mathbf{=}{\mathit{e}}^{\mathbf{t}} \\ \frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{-}\mathit{x}\mathbf{+}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{5}{\mathit{e}}^{\mathbf{t}} \end{gathered}$$

Short answer

The solution is,

$x=c_1{e}^{4t}+\frac{4}{3}{e}^t$

$y=-\frac{3c_1}{4}{e}^{4t}+5e^t+c$

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).

Using Elimination.

Define $\frac{d}{dt}=D$and substitute in the given equations.

$$\begin{gathered} 2Dx-5x+Dy=e^t \\ Dx-x+Dy=5e^t \\ (2D-5)x+Dy=e^t \\ (D-1)x+Dy=5e^t \end{gathered}$$

Eliminate $Dy$:

$$\begin{gathered} (D-1-2D+5)x=4e^t \\ (4-D)x=4e^t.............\left(1\right) \end{gathered}$$

Now we have to obtain $x$using the method of undetermined coefficients:

$$\begin{gathered} D=4 \\ {x}_h=c_1{e}^{4t} \\ {x}_p=A{e}^t \\ x{\text{'}}_p=A{e}^t \end{gathered}$$

To find the particular solutions.

Substitute in the equation (1)

You can treat the operator $D$algebraically or you can apply it to differentiate the term its multiplied by,

$$\begin{gathered} 4A{e}^t-DA{e}^t=4e^t \\ 3A{e}^t=4e^t \end{gathered}$$

Simplify,

$A=\frac{4}{3}$

So, the particular solution is,

$x_p=\frac{4}{3}{e}^t$

The general solution for x is,

$x=c_1{e}^{4t}+\frac{4}{3}{e}^t$

To find the solutions.

From the given, $Dy=(1-D)x+5e^t$,

$$\begin{gathered} Dy=(1-D)(c_1{e}^{4t}+\frac{4}{3}{e}^t)+5e^t \\ Dy=c_1{e}^{4t}+\frac{4}{3}{e}^t-D{c}_1{e}^{4t}-D\frac{4}{3}{e}^t+5e^t \end{gathered}$$

Taking integral on both sides,

$$\begin{gathered} \int Dy=\int \left(c_1{e}^{4t}+\frac{4}{3}{e}^t-D{c}_1{e}^{4t}-D\frac{4}{3}{e}^t+5e^t\right)dt \\ y=\frac{c_1}{4}{e}^{4t}+\frac{4}{3}{e}^t-c_1{e}^{4t}-\frac{4}{3}{e}^t+5e^t+c \\ y=-\frac{3c_1}{4}{e}^{4t}+5e^t+c \end{gathered}$$

Therefore, the general solution is,

$x=c_1{e}^{4t}+\frac{4}{3}{e}^t$

$y=-\frac{3c_1}{4}{e}^{4t}+5e^t+c$