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

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− | + | To solve this system of differential equations, we begin by finding the eigenvalues of the coefficient matrix </br> | |

− | <math> A = | + | :<math> |

+ | A = | ||

+ | \begin{bmatrix} | ||

2 & 1 \\ | 2 & 1 \\ | ||

1 & 2 | 1 & 2 | ||

− | \end{ | + | \end{bmatrix}. |

− | These can be computed by finding the | + | </math> |

− | <math> | + | |

− | 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{ | + | \begin{align} |

− | + | p(\lambda) &= | |

− | + | \det | |

− | This quadratic factors into <math>(\lambda - 3)(\lambda -1)</math> | + | \begin{bmatrix} |

− | We now compute | + | 2-\lambda & 1 \\ |

− | <math> \left | + | 1 & 2-\lambda |

− | + | \end{bmatrix} \\ | |

− | <math> \left | + | &= |

− | + | (2 - \lambda)^2 - 1 \\ | |

− | <math>\begin{align} | + | &= \lambda^2 - 4 \lambda + 4 - 1 \\ |

− | x &= | + | &= \lambda^2 - 4 \lambda + 3 \\ |

− | y &= | + | &= (\lambda - 3)(\lambda - 1). |

− | \end{align}</math | + | \end{align} |

− | Setting t = 0, we | + | </math> |

− | <math>\begin{align} | + | |

− | + | 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>. | |

− | + | ||

− | \end{align}</math | + | We now compute eigenvectors for each eigenvalue. The eigenvectors of eigenvalue 3 solve the equation <math>(A - 3I)\mathbf{v} = \mathbf{0}:</math> |

− | + | ||

− | <math>\begin{align} | + | :<math> \left[\begin{array}{cc|c} -1 & 1 & 0 \\ 1 & -1 & 0 \end{array} \right].</math> |

− | x &= | + | |

− | y &= | + | 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} | + | |

− | + | 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