Difference between revisions of "Tutorial Week 6"

Revision as of 18:08, 9 March 2017

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.

   
   

Solutions

1. (7 pts)The characteristic equation has repeated roots $\lambda=3$ and the two linearly independent solutions are $$\mathbf{x_1}(t)=\mathbf{v_1}e^{3 t},\quad \mathbf{x_2}(t)=(\mathbf{v_1}t+\mathbf{v_2})e^{3 t},$$ where $\mathbf{v_1}$ and $\mathbf{v_2}$ satisfy \begin{equation*} \left( \begin{array}{cc} -1&-1\\ 1&1\\ \end{array} \right)\mathbf{v_1}= \left( \begin{array}{c} 0\\ 0 \end{array} \right) \end{equation*} and \begin{equation*} \left( \begin{array}{cc} -1&-1\\ 1&1\\ \end{array} \right)\mathbf{v_2}= \mathbf{v_1} \end{equation*} One form for the general solution is \begin{equation*} \mathbf{x}(t)=c_1 \left( \begin{array}{c} 1\\ -1 \end{array} \right)e^{3 t}+c_2 \left( \begin{array}{c} t\\ -t-1 \end{array} \right)e^{3 t} \end{equation*} Points: 1 pt for e-value, 1 pt for the e-vector ($\mathbf{v_1}$) equation, 1 pt for calculating the e-vector, 1 pt for the generalized e-vector ($\mathbf{v_2}$) equation, 1 pt for calculating the generalized e-vector ($\mathbf{v_2}$), 1 pt for assembling the $c_1$... part of the solution, 1 pt for assembling the $c_2$... part of the solution.
2. 1. The two eigenvalues are $\lambda=0$ and $\lambda=-3$. The choice of the particular solution is not unique as long as it satisfies \begin{equation*} \left( \begin{array}{cc} 1&-2\\ 2&-4\\ \end{array} \right)\mathbf{x_p}= \left( \begin{array}{c} -1\\ -2 \end{array} \right), \end{equation*} or in other words, \begin{equation*} \left(\begin{array}{c} 2\\ 1 \end{array}\right)s+ \left( \begin{array}{c} -1\\ 0 \end{array} \right). \end{equation*} One of the general solution is \begin{equation*} \mathbf{x}(t)=c_1 \left( \begin{array}{c} 2\\ 1 \end{array} \right)+c_2 \left( \begin{array}{c} 1\\ 2 \end{array} \right)e^{-3 t}+ \left( \begin{array}{c} 1/3\\ 2/3 \end{array} \right) \end{equation*}

   
   

   2. $b_2=2b_1$.
   3. The particular solution has the form $\mathbf{x_p}=t\mathbf{v}+\mathbf{w}.$ The solutions to $\mathbf{A} \mathbf{v} =0$ are $ \mathbf{v}=c \left( \begin{array}{c} 2\\ 1 \end{array} \right) $ and $\mathbf{A} \mathbf{w}=\mathbf{v}-\mathbf{b}$ has to be $d \left( \begin{array}{c} 1\\ 2 \end{array} \right), $ therefore \begin{equation*} \mathbf{v}= \left( \begin{array}{c} 2\\ 1 \end{array} \right),\quad \mathbf{w}= \left( \begin{array}{c} 2\\ 1 \end{array} \right)s+ \left( \begin{array}{c} -1\\ 0 \end{array} \right), \end{equation*} and therefore one form for the general solution is \begin{equation*} \mathbf{x}(t)=c_1 \left( \begin{array}{c} 2\\ 1 \end{array} \right)+c_2 \left( \begin{array}{c} 1\\ 2 \end{array} \right)e^{-3 t}+t \left( \begin{array}{c} 2\\ 1 \end{array} \right) +\left( \begin{array}{c} 1/3\\2/3\end{array}\right),\end{equation*} which is just the one in part (a) plus $\quad t \left(\begin{array}{c}2\\1\end{array}\right).$

   Points:
   (a) 1 pt for characteristic equation, 1 pt for eigenvalues, 2 pts for eigenvectors (1 pt each), 1 pt for a correct particular solution.
   (b) 1 pt for correct b equation.
   (c) 1 pt for getting $\mathbf{v}$, 1 pt for getting $\mathbf{w}$, 1 pt for writing the general solution.