MAT244--2018F > Quiz-5

Q5 TUT 0601

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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= -0.5x_1 + 2x_2, &&x_1(0) = -2,\\ &x'_2= -2x_1 - 0.5x_2, &&x_2(0) = 2 \end{aligned}\right.

Yulin WANG:
(1)
Rewrite the first equation: $x_{2} = \frac{1}{2}x_{1}' + \frac{1}{4}x_{1}$
Then, $x_{2}' = \frac{1}{2}x_{1}'' + \frac{1}{4}x_{1}'$
Plug into the second equation: $\frac{1}{2}x_{1}'' + \frac{1}{4}x_{1}' = -2x_{1} -\frac{1}{2}(\frac{1}{2}x_{1}' + \frac{1}{4}x_{1})$
Then, we get: $x_{1}'' + x_{1}' + \frac{17}{4}x_{1} = 0$ which is  a second order ODE of $𝑥_{1}$
(2)
\begin{align*}
Let A &=
\begin{bmatrix}
-\frac{1}{2} & 2 \\
-2 & -\frac{1}{2}
\end{bmatrix}\\
A-\lambda I &=
\begin{bmatrix}
-\frac{1}{2} -\lambda & 2 \\
-2 & -\frac{1}{2} -\lambda
\end{bmatrix}\\
det(A-\lambda I) &= (\lambda + \frac{1}{2})^{2} + 4 = 0\\
\lambda + \frac{1}{2} &= \pm 2i\\
\lambda &= -\frac{1}{2} \pm 2i\\
\end{align*}
For $\lambda = -\frac{1}{2} + 2i$:
\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^{(-\frac{1}{2} + 2i)t}\begin{bmatrix} 1\\ i \end{bmatrix}\\
&= e^{-\frac{1}{2}t}e^{2it}\begin{bmatrix} 1\\ i \end{bmatrix}\\
&=  e^{-\frac{1}{2}t}(cos2t + isin2t)\begin{bmatrix} 1\\ i \end{bmatrix}\\
&= \begin{bmatrix} cos2t + isin2t\\ icos2t - sin2t \end{bmatrix}\\
Thus, \  \phi_{1}(t) &= e^{-\frac{1}{2}t}\begin{bmatrix} cos2t\\- sin2t \end{bmatrix}\\
\phi_{2}(t) &= e^{-\frac{1}{2}t}\begin{bmatrix} sin2t\\cos2t \end{bmatrix}\\
Therefore, \ x_{1} &= c_{1}e^{-\frac{1}{2}t}cos2t + c_{2}e^{-\frac{1}{2}t}sin2t\\
x_{2} &= -c_{1}e^{-\frac{1}{2}t}sin2t + c_{2}e^{-\frac{1}{2}t}cos2t\\
Since, \ x_{1}(0) &= -2 \ and \  x_{2}(0) = 2\\
So, \ c_{1} &= -2 \ and \ c_{2} = 2\\
\end{align*}
Therefore,
\begin{align*}
x_{1} &= -2e^{-\frac{1}{2}t}cos2t + 2e^{-\frac{1}{2}t}sin2t\\
x_{2} &= 2e^{-\frac{1}{2}t}sin2t + 2e^{-\frac{1}{2}t}cos2t\\
\end{align*}

Monika Dydynski:

--- Quote from: Yulin Wang on November 02, 2018, 03:22:26 PM ---In the attachment.

--- End quote ---

Fyi, your posts are all missing attachments!

Zhiya Lou:
Not sure.... but TUT 0601 actually have this question:

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= -0.5x_1 + 2x_2, &&x_1(0) = -2,\\ &x'_2= -2x_1 - 0.5x_2, &&x_2(0) = 2 \end{aligned}\right.

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

Solving the first equation for $x_2$, we have
$$x_2=\frac{1}{2}x’_1+\frac{1}{4}x_1\tag{1}$$

Plugging $x_2$ into the second equation gives

$$\left(\frac{1}{2}x’_1+\frac{1}{4}x_1\right)’=-2x_1-\frac{1}{2}\left(\frac{1}{2}x’_1+\frac{1}{4}x_1\tag{1}\right)$$
$$x''_1+x'_1+\frac{17}{4}x_1=0$$

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

The characteristic equation is
$$r^2+r+\frac{17}{4}=0$$
$$r_1,2=\frac{1}{2}\pm 2i.$$

The general solution is
$$x_1(t)=e^{-\frac{1}{2}t}\left(c_1\cos{2t}+c_2\sin{2t}\right).$$

It follows that the solution for $x_2$ is

$$x_2(t)=-\frac{1}{2}e^{-\frac{1}{2}t}(c_1\left(-\frac{1}{4}\cos{2t}-2\sin{2t}\right)+c_2\left(2\cos{2t}-\frac{1}{4}\sin{2t}\right))$$

Satisfying the given initial conditions  $x_1(0)=-2$ and $x_2(0)=2$, we have

\left\{\begin{aligned} &c_1=-2\\ &c_2=2 \end{aligned}\right.

The solutions that satisfy the given initial conditions are

$$x_1(t)=e^{-\frac{1}{2}t}(2\sin{2t}-2\cos{2t})$$
$$x_2(t)=e^{-\frac{1}{2}t}(2\cos{2t}+2\sin{2t}).$$