Science:Math Exam Resources/Courses/MATH101/April 2005/Question 04
• Q1 (a) • Q1 (b) • Q1 (c) • Q1 (d) • Q1 (e) • Q1 (f) • Q1 (g) • Q2 (a) • Q2 (b) • Q2 (c) • Q2 (d) • Q3 (a) • Q3 (b) • Q3 (c) • Q4 • Q5 (a) • Q5 (b) • Q6 (a) • Q6 (b) • Q7 (a) • Q7 (b) • Q7 (c) • Q8 (a) • Q8 (b) •
Full-Solution Problem. Justify your answer and show all your work. Simplification of your answer is not required.
A mass attached to a spring satisfies the differential equation
where x = x(t) is the position of the mass at time t.
Find x(t), given that x(0) = x'(0) = 0 (i.e. solve the initial-value problem).
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!
Remember that the solution to the initial value problem will be the sum of a particular solution and a homogeneous solution.
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.
Since this is a non-homogeneous problem, we must find a particular solution and a homogeneous solution. Let and be the particular and homogeneous solutions, resp.
We will first find the particular solution by assuming a oscillatory solution of the same frequency as the right hand side of the differential equation:
where are constants to be determined.
Separating the parts of the equation that are multiplied by and , we find that we must solve the following equations for :
Thus the particular solution is:
Now will find the solution to the homogeneous equation, , which satisfies:
As is standard in homogeneous constant coefficient problems, we substitute into the above equation and get
Factoring and solving for gives
Thus, the homogeneous solution is given by
where the constants , will be chosen to satisfy the initial condition.
Putting both the particular and homogeneous solution together we obtain the full solution:
Now, apply the initial conditions: .
Thus, the solution to the initial value problem is