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Question 08
Consider the linear differential equation ${\begin{aligned}x'(t)&=3x(t)+4y(t)\\y'(t)&=3x(t)+2y(t)\\\end{aligned}}$ with the initial conditions $\displaystyle x(0)=6$, $\displaystyle y(0)=1$. Find the solution $\displaystyle u(t)=(x(t),y(t))^{T}$ and determine the stability of the differential equation.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Start by writing the differential equation in matrix form u' = Au.

Hint 2
If the eigenvalues of the corresponding matrix A are distinct, then the general solution of the linear differential equation is given by $\displaystyle u(t)=c_{1}v_{1}e^{\lambda _{1}t}+c_{2}v_{2}e^{\lambda _{2}t}$ where v1 and v2 are eigenvectors corresponding to the eigenvalues λ1 and λ2, respectively, and c1, c2 are constants.

Hint 3
Next, use the initial condition to solve for the constants c1 and c2.

Hint 4
Finally, the stability is determined by the sign of the (real part of the) eigenvalues.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

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 begin with we rewrite the system in matrix form: ${\begin{bmatrix}x'(t)\\y'(t)\end{bmatrix}}={\begin{bmatrix}3&4\\3&2\end{bmatrix}}{\begin{bmatrix}x(t)\\y(t)\end{bmatrix}}$ and call this matrix A. We need the eigenvalues of A and hence find the zeros of the characteristic polynomials ${\begin{aligned}\det(A-\lambda I)&=\det \left({\begin{bmatrix}3-\lambda &4\\3&2-\lambda \end{bmatrix}}\right)\\&=(3-\lambda )(2-\lambda )-12\\&=\lambda ^{2}-5\lambda -6\\&=(\lambda -6)(\lambda +1)\end{aligned}}$ We see that λ1 = 6 and λ2 = -1. Next, we calculate corresponding eigenvectors. For λ1 = 6 we find the eigenvectors in the null space of ${\begin{bmatrix}-3&4\\3&-4\end{bmatrix}}$ e.g. $v_{1}={\begin{bmatrix}4\\3\end{bmatrix}}$. Similarly, for λ2 = -1 we find the eigenvectors in the null space of ${\begin{bmatrix}4&4\\3&3\end{bmatrix}}$ e.g. $v_{2}={\begin{bmatrix}1\\-1\end{bmatrix}}$. Hence, the general solution of the linear differential equation is given by ${\begin{bmatrix}x(t)\\y(t)\end{bmatrix}}=c_{1}{\begin{bmatrix}4\\3\end{bmatrix}}e^{6t}+c_{2}{\begin{bmatrix}1\\-1\end{bmatrix}}e^{-t}$ To find the values of the constants c1 and c2 we plug in the values at the initial condition t = 0: ${\begin{aligned}x(0)&=6&=4c_{1}+c_{2}\\y(0)&=1&=3c_{1}-c_{2}\end{aligned}}$ which has the solution c1 = 1, c2 = 2. Therefore, the full solution of the linear differential equation with the given initial conditions is $u(t)={\begin{bmatrix}x(t)\\y(t)\end{bmatrix}}={\begin{bmatrix}4\\3\end{bmatrix}}e^{6t}+{\begin{bmatrix}2\\-2\end{bmatrix}}e^{-t}$ Finally, since one of the eigenvalues is positive, the differential equation is unstable.

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