Science:Math Exam Resources/Courses/MATH215/December 2011/Question 06 (c)
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Question 06 (c) 

Let (c) Find the solution to the initial value problem and

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 

From part (b), the homogeneous solution was found. Can you find a particular solution? 
Hint 2 

How would you find the particular solution if this were a scalar first order differential equation? How can you modify that method for a vector solution? 
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 

Found a typo? Is this solution unclear? Let us know here.
Please rate my easiness! It's quick and helps everyone guide their studies. As there are constants and exponentials in this firstorder linear ODE, we will use undetermined coefficients, postulating that the particular solution is
so that
Substituting our solution into the system we get, This must equal and so we have This gives us the system of equations
(based on the terms) and
(based on comparing the constant vectors). Looking at the first equation of the first system of equations, we have so that in the second equation, Using the first equation again, we get Looking at the first equation of the second system of equations, we have . From the second equation we get and hence after substituting back into the first equation. The general solution to the ODE system will be where is our homogeneous solution from part (b). Thus, If then The first component tells us that while using this value of b in the second component gives . Therefore, the solution is
