Difference between revisions of "Tutorial Week 6"

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<!-- See history of "Tutorial Week 7" for solutions -->
 
<!-- See history of "Tutorial Week 7" for solutions -->
  
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=== Solutions ===
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[[Tutorial Week 6 Solutions]]
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<ol>
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<li> ('''7 pts''')The characteristic equation has repeated roots $\lambda=3$ and the two linearly independent solutions are
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$$\mathbf{x_1}(t)=\mathbf{v_1}e^{3 t},\quad \mathbf{x_2}(t)=(\mathbf{v_1}t+\mathbf{v_2})e^{3 t},$$
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where $\mathbf{v_1}$ and $\mathbf{v_2}$ satisfy
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\begin{equation*}
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\left(
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\begin{array}{cc}
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-1&-1\\
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1&1\\
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\end{array}
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\right)\mathbf{v_1}=
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\left(
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\begin{array}{c}
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0\\
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0
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\end{array}
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\right)
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\end{equation*}
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and
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\begin{equation*}
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\left(
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\begin{array}{cc}
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-1&-1\\
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1&1\\
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\end{array}
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\right)\mathbf{v_2}=
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\mathbf{v_1}
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\end{equation*}
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One form for the general solution is
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\begin{equation*}
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\mathbf{x}(t)=c_1
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\left(
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\begin{array}{c}
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1\\
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-1
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\end{array}
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\right)e^{3 t}+c_2
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\left(
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\begin{array}{c}
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t\\
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-t-1
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\end{array}
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\right)e^{3 t}
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\end{equation*}
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Points:  '''1 pt''' for e-value,
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'''1 pt''' for the e-vector ($\mathbf{v_1}$) equation, '''1 pt''' for calculating the e-vector,
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'''1 pt''' for the generalized e-vector ($\mathbf{v_2}$) equation, '''1 pt''' for calculating the generalized e-vector ($\mathbf{v_2}$),
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'''1 pt''' for assembling the $c_1$... part of the solution, '''1 pt''' for assembling the $c_2$... part of the solution.
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</li>
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<li>
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<ol>
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<li>
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The two eigenvalues are $\lambda=0$ and $\lambda=-3$. The choice of the particular solution is not unique as long as it satisfies
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\begin{equation*}
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\left(
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\begin{array}{cc}
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1&-2\\
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2&-4\\
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\end{array}
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\right)\mathbf{x_p}=
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\left(
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\begin{array}{c}
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-1\\
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-2
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\end{array}
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\right),
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\end{equation*}
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or in other words,
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\begin{equation*}
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\left(\begin{array}{c}
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2\\
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1
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\end{array}\right)s+
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\left(
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\begin{array}{c}
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-1\\
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0
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\end{array}
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\right).
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\end{equation*}
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One of the general solution is
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\begin{equation*}
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\mathbf{x}(t)=c_1
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\left(
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\begin{array}{c}
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2\\
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1
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\end{array}
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\right)+c_2
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\left(
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\begin{array}{c}
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1\\
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2
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\end{array}
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\right)e^{-3 t}+
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\left(
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\begin{array}{c}
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1/3\\
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2/3
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\end{array}
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\right)
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\end{equation*}</li>
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<li> $b_2=2b_1$. </li>
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<li>The particular solution has the form
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$\mathbf{x_p}=t\mathbf{v}+\mathbf{w}.$
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The solutions to $\mathbf{A} \mathbf{v} =0$ are
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$
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\mathbf{v}=c
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\left(
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\begin{array}{c}
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2\\
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1
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\end{array}
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\right)
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$
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and $\mathbf{A} \mathbf{w}=\mathbf{v}-\mathbf{b}$ has to be $d
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\left(
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\begin{array}{c}
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1\\
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2
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\end{array}
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\right),
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$
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therefore \begin{equation*}
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\mathbf{v}=
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\left(
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\begin{array}{c}
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2\\
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1
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\end{array}
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\right),\quad
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\mathbf{w}=
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\left(
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\begin{array}{c}
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2\\
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1
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\end{array}
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\right)s+
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\left(
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\begin{array}{c}
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-1\\
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0
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\end{array}
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\right),
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\end{equation*}
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and therefore one form for the general solution is
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\begin{equation*}
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\mathbf{x}(t)=c_1
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\left(
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\begin{array}{c}
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2\\
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1
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\end{array}
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\right)+c_2
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\left(
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\begin{array}{c}
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1\\
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2
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\end{array}
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\right)e^{-3 t}+t
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\left(
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\begin{array}{c}
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2\\
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1
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\end{array}
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\right)
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+\left(
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\begin{array}{c}
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1/3\\2/3\end{array}\right),\end{equation*}
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which is just the one in part (a) plus
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$\quad t \left(\begin{array}{c}2\\1\end{array}\right).$
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</li>
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</ol>
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Points:
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(a) '''1 pt''' for characteristic equation, '''1 pt''' for eigenvalues, '''2 pts''' for eigenvectors (1 pt each), '''1 pt''' for a correct particular solution.
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(b) '''1 pt''' for correct b equation.
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(c) '''1 pt''' for getting $\mathbf{v}$, '''1 pt''' for getting $\mathbf{w}$, '''1 pt''' for writing the general solution.
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</li>
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</ol>
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Latest revision as of 23:06, 30 December 2020

Worksheet Questions

You can print the PDF. This worksheet relates to material from weeks 6 and 7.

  1. Find the general solution to the following linear system of differential equations \begin{equation} \mathbf{x}'= \left( \begin{array}{cc} 2&-1\\ 1&4 \end{array} \right)\mathbf{x} \end{equation}


  2. Consider the following system \begin{equation*} \mathbf{x}'= \left( \begin{array}{cc} 1&-2\\ 2&-4 \end{array} \right)\mathbf{x}+ \mathbf{b},\quad \text{where} \quad \mathbf{b}= \left( \begin{array}{c} b_1\\ b_2 \end{array} \right) \end{equation*}
    1. Determine the general solution when $b_1=1$ and $b_2=2$.
    2. In part (a), $b_1$ and $b_2$ were specially chosen in that the row reduced form of the matrix equation had a full row of zeros (including the RHS) and therefore a solution. Write down an equation for $b_1$ and $b_2$ that ensures this will happen.
    3. Now consider the case with $b_1=3$ and $b_2=3$. Because this $\mathbf{b}$ does not satisfy the equation from part (b), a different form for your particular guess is needed. By analogy with second order systems, we guess $\mathbf{x_p} = t \mathbf{v}+\mathbf{w}$. Now we take the following steps to find out the general solution for this case:
      1. Plugging this $\mathbf{x_p} $ into the system of ODEs, we find that we must have $\mathbf{v} = t \mathbf{A} \mathbf{v}+ \mathbf{A} \mathbf{w} + \mathbf{b}$ for all $t$. This requires that $\mathbf{A} \mathbf{v}=0$ and $\mathbf{A} \mathbf{w}=\mathbf{v}-\mathbf{b};$
      2. $\mathbf{A} \mathbf{v}=0$ has a whole family of solutions. In fact , since $0$ is an eigenvalue of $\mathbf{A} $, $\mathbf{v}$ should be a corresponding eigenvector;
      3. Next consider the equation $\mathbf{A} \mathbf{w}=\mathbf{v}-\mathbf{b}$ which we must solve for $\mathbf{w}$. Notice that for any vector $\mathbf{w}$, $\mathbf{A} \mathbf{w}$ will always have a second component equal to twice its first component. Thus, to be able to solve $\mathbf{A} \mathbf{w}=\mathbf{v}-\mathbf{b}$ , we must make sure that $\mathbf{v}-\mathbf{b}$ has second component equal to twice its first component. Find the vector $\mathbf{v}$ from the family of solutions to $\mathbf{A} \mathbf{v}=0$ that does this. Then find $\mathbf{w}$ and write down the general solution for this case.



Tutorial Week 6 Solutions

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