Science:Math Exam Resources/Courses/MATH307/December 2008/Question 08
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Question 08 

Consider the linear differential equation with the initial conditions , . Find the solution 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 where v_{1} and v_{2} are eigenvectors corresponding to the eigenvalues λ_{1} and λ_{2}, respectively, and c_{1}, c_{2} are constants. 
Hint 3 

Next, use the initial condition to solve for the constants c_{1} and c_{2}. 
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.

Solution 

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Please rate my easiness! It's quick and helps everyone guide their studies. To begin with we rewrite the system in matrix form: and call this matrix A. We need the eigenvalues of A and hence find the zeros of the characteristic polynomials We see that λ_{1} = 6 and λ_{2} = 1. Next, we calculate corresponding eigenvectors. For λ_{1} = 6 we find the eigenvectors in the null space of e.g. . Similarly, for λ_{2} = 1 we find the eigenvectors in the null space of e.g. . Hence, the general solution of the linear differential equation is given by To find the values of the constants c_{1} and c_{2} we plug in the values at the initial condition t = 0: which has the solution c_{1} = 1, c_{2} = 2. Therefore, the full solution of the linear differential equation with the given initial conditions is Finally, since one of the eigenvalues is positive, the differential equation is unstable. 