MAT244--2018F > Quiz-5

Q5 TUT 5101

(1/1)

Victor Ivrii:
Transform the given system into a single equation of second order and find the solution $(x_1(t),x_2(t))$, satisfying initial conditions
\left\{\begin{aligned} &x'_1= 2x_2, &&x_1(0) = 3,\\ &x'_2= -2x_1, &&x_2(0) = 4. \end{aligned}\right.

Yulin WANG:
(1)
Rewrite the first equation: $x_{2} = \frac{1}{2}x_{1}'$
Then: $x_{2}'' = \frac{1}{2}x_{1}''$
Plug into the second equation: $\frac{1}{2}x_{1}'' = -2x_{1}$
Therefore, we get $x_{1}'' + 4x_{1} = 0$ which is a second order ODE of $𝑥_{1}$
(2)
\begin{align*}
Let A &=
\begin{bmatrix}
0 & 2 \\
-2 & 0
\end{bmatrix}\\
A-\lambda I &=
\begin{bmatrix}
-\lambda & 2 \\
-2 & -\lambda
\end{bmatrix}\\
det(A-\lambda I) &= \lambda^{2} + 4 = 0\\
\lambda &= \pm 2i\\
\end{align*}
For $\lambda = 2i$:
\begin{align*}
A-\lambda I &=
\begin{bmatrix}
-2i & 2 \\
-2 & -2i
\end{bmatrix}\\
null\begin{bmatrix} -2i & 2 \\ -2 & -2i \end{bmatrix}
&= span{\begin{bmatrix} 1\\ i \end{bmatrix}\\}\\
so, \ the \ eigenvector \  V &= \begin{bmatrix} 1\\ i \end{bmatrix}\\
Then, \  e^{\lambda t}V &= e^{2it}\begin{bmatrix} 1\\ i \end{bmatrix}\\
&= (cos2t + isin2t)\begin{bmatrix} 1\\ i \end{bmatrix}\\
&= \begin{bmatrix} cos2t + isin2t\\ icos2t - sin2t \end{bmatrix}\\
Thus, \  \phi_{1}(t) &= \begin{bmatrix} cos2t\\- sin2t \end{bmatrix}\\
\phi_{2}(t) &= \begin{bmatrix} sin2t\\cos2t \end{bmatrix}\\
Therefore, \ x_{1} &= c_{1}cos2t + c_{2}sin2t\\
x_{2} &= -c_{1}sin2t + c_{2}cos2t\\
Since, \ x_{1}(0) &= 3 \ and \  x_{2}(0) = 4\\
So, \ c_{1} &= 3 \ and \ c_{2} = 4\\
\end{align*}
Therefore,
\begin{align*}
x_{1} &= 3cos2t + 4sin2t\\
x_{2} &= -3sin2t + 4cos2t\\
\end{align*}

Monika Dydynski:
Transform the given system into a single equation of second order and find the solution (x_1(t), x_2(t)), satisfying initial conditions
\left\{\begin{aligned} &x'_1= 2x_2, &&x_1(0) = 3,\\ &x'_2= -2x_1, &&x_2(0) = 4. \end{aligned}\right.

a. Transform the given system into a single equation of second order.

Solving the first equation for $x_2$, we have
$$x_2=\frac{1}{2}x'_1$$

Plugging $x_2$ into the second equation gives

$$\left(\frac{1}{2}x’_1\right)’=-2x_1$$
$$\frac{1}{2}x’’_1+2x’_1=0$$
$$x’’_1+4x’_1=0$$

b. Find $x_1$ and $x_2$ that also satisfy the given initial conditions.

The characteristic equation is
$$\frac{1}{2}r^2+2=0$$
$$r^2+4=0$$
$$r_1,2=\pm 2i$$

The general solution is
$$x_1(t)=c_1\cos{2t}+c_2\sin{2t}.$$

It follows that the solution for $x_2$ is

$$x_2(t)=\frac{1}{2}(-2c_1\sin{2t}+2c_2\cos{2t})=-c_1\sin{2t}+c_2 \cos{2t}$$

Satisfying the given initial conditions  $x_1(0)=3$ and $x_2(0)=4$, we have

\left\{\begin{aligned} &c_1=3\\ &c_2=4\end{aligned}\right.

The solutions that satisfy the given initial conditions are

$$x_1(t)=3\cos{2t}+4\sin{2t}$$
$$x_2(t)=-3\sin{2t}+4\cos{2t}.$$

Chuqiao Liu:
Answer in the attachment.

Victor Ivrii:
Yulin, you are asked to solve this problem in a very specific way. Still in the combination with Monika's post it is a useful demo.

Also \det , \cos . ...

Chuqiao
You posted solution to some other problem