Science:Math Exam Resources/Courses/MATH152/April 2015/Question A 27/Solution 1

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

$A={\begin{bmatrix}2&1\\1&2\end{bmatrix}}.$

These can be computed by finding the roots of $p(\lambda )=\det(A-\lambda I):$

${\begin{aligned}p(\lambda )&=\det {\begin{bmatrix}2-\lambda &1\\1&2-\lambda \end{bmatrix}}\\&=(2-\lambda )^{2}-1\\&=\lambda ^{2}-4\lambda +4-1\\&=\lambda ^{2}-4\lambda +3\\&=(\lambda -3)(\lambda -1).\end{aligned}}$

This quadratic factors into $(\lambda -3)(\lambda -1)$, so the eigenvalues of the coefficient matrix are $\lambda _{1}=3,\lambda _{2}=1$.

We now compute eigenvectors for each eigenvalue. The eigenvectors of eigenvalue 3 solve the equation $(A-3I)\mathbf {v} =\mathbf {0} :$

$\left[{\begin{array}{cc|c}-1&1&0\\1&-1&0\end{array}}\right].$

Writing $\mathbf {v} =(v_{1},v_{2})$, these equations imply that $-v_{1}+v_{2}=0$, and hence $v_{1}=v_{2}$. Thus $\mathbf {v} _{1}=(1,1)$ is an eigenvector of eigenvalue 3.

Similarly, for the eigenvalue 1, we have the system

$\left[{\begin{array}{cc|c}1&1&0\\1&1&0\end{array}}\right],$

so $v_{1}=-v_{2}$. Hence $\mathbf {v} _{2}=(-1,1)$ is an eigenvector of eigenvalue 1.

The general form of the solution is therefore

${\begin{aligned}x(t)&=c_{1}e^{3t}-c_{2}e^{t}\\y(t)&=c_{1}e^{3t}+c_{2}e^{t}.\end{aligned}}$

Setting $t=0$ and using the initial conditions, we obtain the equations

${\begin{aligned}c_{1}-c_{2}&=2\\c_{1}+c_{2}&=3,\end{aligned}}$

whence $c_{1}={\tfrac {5}{2}},c_{2}={\tfrac {1}{2}}$. Thus the solution is

$\color {blue}{\begin{aligned}x(t)&={\tfrac {5}{2}}e^{3t}-{\tfrac {1}{2}}e^{t}\\y(t)&={\tfrac {5}{2}}e^{3t}+{\tfrac {1}{2}}e^{t}.\end{aligned}}$