Question

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

$$\begin{gathered} \frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{-}\mathit{y}\mathbf{+}\mathit{t} \\ \frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{x}\mathbf{-}\mathit{t} \end{gathered}$$

Short answer

The solution is,

$x\left(t\right)=c_1\cos t+c_2\sin t+t+1$and $y\left(t\right)=c_1\sin t-c_2\cos t+t-1$

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.

We have the system of first order equations

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

Which is

$$\begin{gathered} y=t-x\text{'} \\ y\text{'}=x-t \end{gathered}$$

Which by using differential operating is equivalent to

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

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

Operate the first equation by the operator D multiply the second equation by -1, then we obtain

$Dy=Dt-D^{2}x$.........(3)

$-Dy=-x+t$............(4)

After that add equation (3) to equation (4) and simplify then we obtain

$$\begin{gathered} Dy-Dy=Dt-D^{2}x-x+t \\ 0=1-D^{2}x-x+t \\ {D}^{2}x+x=t+1 \end{gathered}$$

$(D^2+1)x=t+1$...........(5)

After that, we have to obtain the solution of the corresponding homogeneous equation $(D^2+1)x=0$as the following:

We have the auxiliary solution for $x$as

$(m^2+1)=0$

Which has the roots

$m_{1,2}=\pm i$

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

$x_c=c_1\cos t+c_2\sin t$............(6)

To find the solutions.

After that, we have to obtain its particular solution by assuming that $x_p=At+B$, then differentiate with respect to $t$is

$$\begin{gathered} D{x}_p=A \\ {D}^2{x}_p=0 \end{gathered}$$

Then substitute with $x_p=At+B$and the second derivative into equation (5), then we obtain

$$\begin{gathered} 0+At+B=t+1 \\ At+B=t+1 \end{gathered}$$

Then by comparison, we can obtain

$$\begin{gathered} A=1 \\ B=1 \end{gathered}$$

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

$x_p=t+1$...........(7)

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

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

$x\left(t\right)=c_1\cos t+c_2\sin t+t+1$............(8)

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

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

$Dx=-c_1\sin t+c_2\cos t+1$.............(9)

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

$\begin{array}{rcl}y\left(t\right)& =& t-Dx\\ & =& t-(-c_1\sin t+c_2\cos t+1)\\ & =& c_1\sin t-c_2\cos t+t-1\end{array}$

$y\left(t\right)=c_1\sin t-c_2\cos t+t-1$is the general solution for $y$.

Therefore, $x\left(t\right)=c_1\cos t+c_2\sin t+t+1$and $y\left(t\right)=c_1\sin t-c_2\cos t+t-1$.