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

MATH152 April 2015

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Question A 27
Find the solution of the system of differential equations ${\begin{aligned}x'&=2x+y\\y'&=x+2y\end{aligned}}$ that satisfies the initial conditions $x(0)=2$ and $y(0)=3$.

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If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Compute the eigenvalues and eigenvectors of the coefficient matrix $A={\begin{bmatrix}2&1\\1&2\end{bmatrix}}.$ If $A$ has two eigenvalues $\lambda _{1},\lambda _{2}$ with associated eigenvectors $\mathbf {v} _{1},\mathbf {v} _{2}$, then ${\begin{bmatrix}x\\y\end{bmatrix}}=c_{1}e^{\lambda _{1}t}\mathbf {v} _{1}+c_{2}e^{\lambda _{2}t}\mathbf {v} _{2}$ for some constants $c_{1},c_{2}$.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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 $\color {blue}{\begin{aligned}x(t)&={\tfrac {5}{2}}e^{3t}-{\tfrac {1}{2}}e^{t}\\y(t)&={\tfrac {5}{2}}e^{3t}+{\tfrac {1}{2}}e^{t}.\end{aligned}}$