Question

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

$$\begin{gathered} 14.\frac{dx}{dt}+\frac{dy}{dt}=e^t \\ -\frac{d^{2}x}{d{t}^2}+\frac{dy}{dt}+x+y=0 \end{gathered}$$

Short answer

The solution is,

$x\left(t\right)=c_1+c_{2}t+c_3{e}^t+t{e}^t$ and $y\left(t\right)=-(c_1+c_2)-c_{2}t-(c_3-1)e^t-t{e}^t$.

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).

Using Elimination.

We have the system of first order equations

$$\begin{gathered} \frac{dx}{dt}+\frac{dy}{dt}=e^t \\ -\frac{d^{2}x}{d{t}^2}+\frac{dx}{dt}+x+y=0 \end{gathered}$$

Which is

$$\begin{gathered} x\text{'}+y\text{'}=e^t \\ -x\text{'}\text{'}+x\text{'}+x+y=0 \end{gathered}$$

Which by using differential operating is equivalent to

$$\begin{gathered} Dx+Dy=e^t........\left(1\right) \\ -(D^2-D-1)x+y=0.........\left(2\right) \end{gathered}$$

And we have to solve it using systematic elimination as the following technique:

Operate the second equation by the operator and multiply the first equation by , then we obtain

$$\begin{gathered} -Dx-Dy=-e^t........\left(1a\right) \\ -D(D^2-D-1)x+Dy=0.........\left(2a\right) \end{gathered}$$

After that add equation (1a) to equation (2a) and simplify then we obtain

$$\begin{gathered} -Dx-Dy-D(D^2-D-1)x+Dy=-e^t \\ [-\mathrm{Dx}-D(D^2-D-1\left)x\right]+[Dy-Dy]=-e^t \\ (-D^3+D^2)x=-e^t \\ (D^2-D^3)x=-e^t..................\left(3\right) \end{gathered}$$

After that, we have to obtain the solution of the corresponding homogeneous equation $(D^3-D^2)x=0$as the following:

We have the auxiliary solution forrole="math" localid="1663958748177" $x$as

$$\begin{gathered} (m^3-m^2)=0 \\ {m}^2(m-1)=0 \end{gathered}$$

Then we have the roots

$m_{1,2}=0$and $m_3=1$

Then we can obtain the homogeneous solution for $x$as

$x_c=c_1+c_{2}t+c_3{e}^t$............(4)

To find the solutions.

After that, we have to obtain its particular solution by assuming that $x_p=At{e}^t$,where we have $e^t$as a solution in the homogeneous part,then differentiate with respect to $t$as

$$\begin{gathered} D{x}_p=A(t+1)e^t \\ {D}^2{x}_p=A(t+2)e^t............\left(5\right) \\ {D}^3{x}_p=A(t+3)e^t...............\left(6\right) \end{gathered}$$

Then substitute with equation (5) and (6) into (3), then we obtain

$$\begin{gathered} A(t+3)e^t-A(t+2)e^t=e^t \\ A{e}^t=e^t \end{gathered}$$

Then we can obtain

$A=1$

Now we can obtain the particular solution for $x$as,

$x_p=t{e}^t$...........(7)

Then from equation (5) and (7), we can obtain the general solution for $x$as,

$x\left(t\right)=x_c+x_p$

$x\left(t\right)=c_1+c_{2}t+c_3{e}^t+t{e}^t$............(8)

Now we have to obtain the general solution for $y$as the following:

Differentiate the solution of $x$shown in equation (8) as

$$\begin{gathered} Dx=c_2+c_3{e}^t+t{e}^t+e^t............\left(9\right) \\ {D}^{2}x=c_3{e}^t+t{e}^t+2e^t...............\left(10\right) \end{gathered}$$

After that, substitute with equation (8), (9), (10) into equation (2), then we obtain

$$\begin{gathered} -(c_3{e}^t+t{e}^t+2e^t)+(c_2+c_3{e}^t+t{e}^t+e^t)+(c_1+c_{2}t+c_3{e}^t+t{e}^t)+y=0 \\ (c_1+c_2)+c_{2}t+(c_3-1)e^t+t{e}^t+y=0 \end{gathered}$$

Then we obtain,

$y\left(t\right)=-(c_1+c_2)-c_{2}t-(c_3-1)e^t-t{e}^t$is the general solution for $y$.

Therefore,$x\left(t\right)=c_1+c_{2}t+c_3{e}^t+t{e}^t$ and$y\left(t\right)=-(c_1+c_2)-c_{2}t-(c_3-1)e^t-t{e}^t$ .