Difference between revisions of "Science:Math Exam Resources/Courses/MATH152/April 2015/Question A 27/Solution 1"

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In order to solve this system of differential equations, we begin by finding the eigenvalues of the coefficient matrix </br>
+
To solve this system of differential equations, we begin by finding the eigenvalues of the coefficient matrix </br>
<math> A = \left( \begin{array}{cc}
+
:<math>  
 +
A =
 +
\begin{bmatrix}
 
2 & 1 \\
 
2 & 1 \\
 
1 & 2  
 
1 & 2  
\end{array} \right)</math>
+
\end{bmatrix}.
These can be computed by finding the determinant of <math>A - \lambda I</math>:</br>
+
</math>
<math> \left| \begin{array}{cc}
+
 
2 - \lambda & 1 \\
+
These can be computed by finding the roots of <math>p(\lambda) = \det(A - \lambda I):</math>
1 & 2 - \lambda
+
:<math>  
\end{array} \right|</math></br>
+
\begin{align}
This is </br>
+
p(\lambda) &=
<math> (2 - \lambda)^2 - 1 = \lambda^2 - 4 \lambda + 4 - 1 = \lambda^2 - 4 \lambda + 3</math></br>
+
\det
This quadratic factors into <math>(\lambda - 3)(\lambda -1)</math>. So the eigenvalues of the system are 3 and 1.</br>
+
\begin{bmatrix}
We now compute the eigenvectors associated to each eigenvalue. The eigenvector v associated to 3 solves the equation (A - 3I)v = 0:</br>
+
2-\lambda & 1 \\
<math> \left(\begin{array}{cc} -1 & 1 \\ 1 & -1 \end{array} \right) \left(\begin{array}{c} x \\ y \end{array} \right) = \left(\begin{array}{c} 0 \\ 0 \end{array} \right) </math></br>
+
1 & 2-\lambda  
Thus -x +y = 0, and x = y. So (1,1) is an eigenvector associated to this eigenvalue. Similarly, the eigenvector associated to 1 solves</br>
+
\end{bmatrix} \\
<math> \left(\begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array} \right) \left(\begin{array}{c} x \\ y \end{array} \right) = \left(\begin{array}{c} 0 \\ 0 \end{array} \right) </math></br>
+
&=
So x = -y. This corresponds to an eigenvector of (-1, 1). So we get the general form for the solution:</br>
+
(2 - \lambda)^2 - 1 \\
<math>\begin{align}
+
&= \lambda^2 - 4 \lambda + 4 - 1 \\
x &= ae^{3t} - be^t \\
+
&= \lambda^2 - 4 \lambda + 3 \\
y &= ae^{3t} + be^t
+
&= (\lambda - 3)(\lambda - 1).
\end{align}</math></br>
+
\end{align}
Setting t = 0, we get the equations:</br>
+
</math>
<math>\begin{align}
+
 
a - b & = 2 \\
+
This quadratic factors into <math>(\lambda - 3)(\lambda -1)</math>, so the eigenvalues of the coefficient matrix are <math>\lambda_1 = 3, \lambda_2 = 1</math>.
a + b & = 3
+
 
\end{align}</math></br>
+
We now compute eigenvectors for each eigenvalue. The eigenvectors of eigenvalue 3 solve the equation <math>(A - 3I)\mathbf{v} = \mathbf{0}:</math>
This has the solution <math> a = 5/2, b = 1/2</math> So we get the system</br>
+
 
<math>\begin{align}
+
:<math> \left[\begin{array}{cc|c} -1 & 1 & 0 \\ 1 & -1 & 0 \end{array} \right].</math>
x &= 5/2 e^{3t} - 1/2 e^t \\
+
 
y &= 5/2 e^{3t} + 1/2 e^t
+
Writing <math>\mathbf{v} = (v_1, v_2)</math>, these equations imply that <math>-v_1 + v_2 = 0</math>, and hence <math>v_1 = v_2</math>. Thus <math>\mathbf{v}_1 = (1,1)</math> is an eigenvector of eigenvalue 3.  
\end{align}</math></br></br>
+
 
'''Answer:''' <math>\color{blue} x = 5/2 e^{3t} - 1/2 e^t ; y = 5/2 e^{3t} + 1/2 e^t</math>
+
Similarly, for the eigenvalue 1, we have the system
 +
 
 +
:<math> \left[\begin{array}{cc|c} 1 & 1 & 0 \\ 1 & 1 & 0 \end{array} \right],</math>
 +
 
 +
so <math>v_1 = -v_2</math>. Hence <math>\mathbf{v}_2 = (-1, 1)</math> is an eigenvector of eigenvalue 1.  
 +
 
 +
The general form of the solution is therefore
 +
 
 +
:<math>\begin{align}
 +
x(t) &= c_1 e^{3t} - c_2 e^t \\
 +
y(t) &= c_1 e^{3t} + c_2 e^t.
 +
\end{align}
 +
</math>
 +
 
 +
Setting <math>t = 0</math> and using the initial conditions, we obtain the equations
 +
 
 +
:<math>\begin{align}
 +
c_1 - c_2 &= 2 \\
 +
c_1 + c_2 &= 3,
 +
\end{align}</math>
 +
 
 +
whence <math> c_1 = \tfrac52, c_2 = \tfrac12</math>. Thus the solution is
 +
 
 +
:<math>
 +
\color{blue}
 +
\begin{align}
 +
x(t) &= \tfrac52 e^{3t} - \tfrac12 e^t \\
 +
y(t) &= \tfrac52 e^{3t} + \tfrac12 e^t.
 +
\end{align}</math>

Latest revision as of 18:27, 8 March 2018

To solve this system of differential equations, we begin by finding the eigenvalues of the coefficient matrix

These can be computed by finding the roots of

This quadratic factors into , so the eigenvalues of the coefficient matrix are .

We now compute eigenvectors for each eigenvalue. The eigenvectors of eigenvalue 3 solve the equation

Writing , these equations imply that , and hence . Thus is an eigenvector of eigenvalue 3.

Similarly, for the eigenvalue 1, we have the system

so . Hence is an eigenvector of eigenvalue 1.

The general form of the solution is therefore

Setting and using the initial conditions, we obtain the equations

whence . Thus the solution is