Question

In Problems 21 and 22 solve the given initial - value problem.

22.

$$\begin{gathered} \frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{y}\mathbf{-}\mathbf{1} \\ \frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{-}\mathbf{3}\mathit{x}\mathbf{+}\mathbf{2}\mathit{y} \\ \mathit{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0} \end{gathered}$$

Short answer

$$\begin{gathered} x\left(t\right)=-\frac{2}{3}\left(e^t\cos \sqrt{2}t+\frac{\sqrt{2}}{4}{e}^t\sin \sqrt{2}t-1\right) \\ y\left(t\right)=-e^t\cos \sqrt{2}t+\frac{1}{\sqrt{2}}{e}^t\sin \sqrt{2}t+1 \end{gathered}$$

Step-by-step solution

To Find the system of first order equations

We have the system of first order equations

$$\begin{gathered} \frac{dx}{dt}=y-1 \\ \frac{dy}{dt}=-3x+2y \end{gathered}$$

with the initial conditions

$x\left(0\right)=0\text{and}y\left(0\right)=0$

which by using differential operating are equivalent to

$$\begin{gathered} Dx-y=-1 \\ Dy-2y+3x=0 \\ (D-2)y+3x=0 \end{gathered}$$

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

Operate the first equation by the operator (D-2), then we obtain

$D(D-2)x-(D-2)y=(D-2)(-1)$

After that, add equation (3) to equation (2) and simplify, then we obtain

$$\begin{gathered} \left[D\right(D-2)x-(D-2\left)y\right]+\left[\right(D-2)y+3x]=(D-2)(-1) \\ \left[\right(D-2)y-(D-2\left)y\right]+\left[D\right(D-2)x+3x]=D(-1)+2 \\ D(D-2)x+3x=2 \\ \left(D^2-2D+3\right)x=2 \end{gathered}$$

Differential equation

Now, we have to obtain the solution of this differential equation as the following technique:

First, we have to find the homogeneous of the corresponding homogeneous equation$\left(D^2-2D+1\right)x=0$ by assuming that$x=e^{mt}$ is a solution for this D.E, then we can have the auxiliary equation as

$m^2-2m+3=0$

which has the roots

$m_{1,2}=\frac{2\pm \sqrt{4-4\times 1\times 3}}{2}=\frac{2\pm \sqrt{-8}}{2}=1\pm \sqrt{2}i$

Then we obtain

$x_h=c_1{e}^t\cos \sqrt{2}t+c_2{e}^t\sin \sqrt{2}t$

is the general solution for x.

Second, we have to find the particular solution by assuming that $x_p=A$, then differentiate with respect to x as

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

After that, substitute with$x_p=A$ and the first and second derivatives into equation (5), then we have

$$\begin{gathered} 0-2\left(0\right)+3A=2 \\ A=\frac{2}{3} \end{gathered}$$

Then we have the particular solution as

$x_p=\frac{2}{3}$

Then from equations (5) and (6), we obtain

$$\begin{gathered} x\left(t\right)=x_h+x_p \\ =c_1{e}^t\cos \sqrt{2}t+c_2{e}^t\sin \sqrt{2}t+\frac{2}{3} \end{gathered}$$

Now we have to obtain the general solution for by substituting with the solution of shown in equation (6) into equation (1) as

$$\begin{gathered} y=Dx+1 \\ =D\left(c_1{e}^t\cos \sqrt{2}t+c_2{e}^t\sin \sqrt{2}t+\frac{2}{3}\right)+1 \\ =\left(c_1{e}^t\cos \sqrt{2}t-\sqrt{2}{c}_1{e}^t\sin \sqrt{2}t+c_2{e}^t\sin \sqrt{2}t+\sqrt{2}{c}_2{e}^t\cos \sqrt{2}t\right)+1 \\ =\left(c_1+\sqrt{2}{c}_2\right)e^t\cos \sqrt{2}t+\left(c_2-\sqrt{2}{c}_1\right)e^t\sin \sqrt{2}t+1 \end{gathered}$$

The initial conditions

Now we have to apply the initial conditions on the solutions of x and y as the following :

First, apply the point$(x,t)=(0,0)$ into equation (7), then we have

$$\begin{gathered} 0=c_1{e}^0\cos \left(0\right)+c_2{e}^0\sin \left(0\right)+\frac{2}{3} \\ 0=c_1+\frac{2}{3} \\ {c}_1=-\frac{2}{3} \end{gathered}$$

Second, apply the point$(y,t)=(0,0)$ into equation (8), then we have

$$\begin{gathered} 0=\left(c_1+\sqrt{2}{c}_2\right)e^0\cos \left(0\right)+\left(c_2-\sqrt{2}{c}_1\right)e^0\sin \left(0\right)+1 \\ {c}_1+\sqrt{2}{c}_2=-1 \end{gathered}$$

Then by solving the two equations (a) and (b) with each other, we can find that

$c_1=-\frac{2}{3}\text{and}{c}_2=-\frac{1}{3\sqrt{2}}$

Final Answer

Finally, we have to substitute with the value of constants $c_{1}and{c}_2$into equations (6) and (7), then we have

$$\begin{gathered} x\left(t\right)=-\frac{2}{3}{e}^t\cos \sqrt{2}t-\frac{1}{3\sqrt{2}}{e}^t\sin \sqrt{2}t+\frac{2}{3} \\ =-\frac{2}{3}\left(e^t\cos \sqrt{2}t+\frac{\sqrt{2}}{4}{e}^t\sin \sqrt{2}t-1\right) \end{gathered}$$

and

$$\begin{gathered} y\left(t\right)=\left(-\frac{2}{3}-\frac{1}{3}\right)e^t\cos \sqrt{2}t+\left(-\frac{1}{3\sqrt{2}}+\frac{2\sqrt{2}}{3}\right)e^t\sin \sqrt{2}t+1 \\ =-e^t\cos \sqrt{2}t+\frac{1}{\sqrt{2}}{e}^t\sin \sqrt{2}t+1 \end{gathered}$$