Science:Math Exam Resources/Courses/MATH105/April 2013/Question 03 (a)
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Question 03 (a) 

Solve the following 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 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 

What kind of differential equation is given in this initial value problem? Which method should you use to solve it? 
Hint 2 

This is a separable differential equation. Isolate the terms with y and with x on separate sides. Then integrate. 
Hint 3 

For the more complicated integral, use partial fractions. 
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. First, we recognize that the differential equation given in this initial value problem is a separable differential equation. Hence, we move all terms involving to opposite sides of the equal sign and take the indefinite integral: The integral on the left hand side is easy, but some work is required to solve the integral on the right hand side. First, we rewrite it as We proceed by using partial fractions to split up the integral. Notice that the fraction can be split up as This gives Setting x to 0 gives and setting x to 1 gives . Thus, we can write out integral as Now we integrate both sides of the equation and add the arbitary constant, , to the right hand side: Isolating for we get: where . Since the initial condition, , we take the positive solution for : To solve for , we apply the initial condition to the solution: Therefore, the solution to the initial value problem is:
