Science:Math Exam Resources/Courses/MATH152/April 2010/Question B 05 (b)
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Question B 05 (b) 

Consider the differential equation system where A has an eigenvalue with corresponding eigenvector Recall that for real matrices, eigenvalues and eigenvectors come in complex conjugate pairs. Find the real form (no imaginary terms) of the solution that satisfies initial conditions 
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 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 

The two particular solutions from part (a) (each was formed using a pair of eigenvalue and eigenvector of A), i.e. come in complex conjugate pairs too. That means they satisfy . Now, the general complexvalued solution from part (a) is a linear combination of these with arbitrary complex constants . What form of such constants, then, must we choose so that when placed in front of and , we can obtain a realvalued solution? You might need to concentrate on the follow fact for complex numbers: is always a real number for any complex number z. Think about how should be related to each other. 
Checking a solution serves two purposes: helping you if, after having used the hint, 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. Continuing from the hint, to choose realvalued solutions, we choose , so that the general solution becomes: Recall that Since The general form of realvalued solutions is: where and are real constants. Their relationship with the previous constants are and Now if we impose the initial condition , we get Thus, and , and the solution of the initial value problem is Alternatively, one can solve directly from And impose the initial condition direcly to find that . Then compute the real part to get the same result as above. The amount of algebra involved will be slightly less. 