Science:Math Exam Resources/Courses/MATH101/April 2006/Question 04 (b)
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Question 04 (b) 

FullSolution Problems. Justify your answers and show all your work. Simplification of answers is not required. Solve the initialvalue problem Solve the initialvalue problem subject to the conditions 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 

This is actually kind of a first order differential equation disguised as a second order differential equation. Try an integrating factor. 
Hint 2 

Multiply both sides of the equation by
where the 2 above came from the coefficient of the term. 
Hint 3 

Even after all this work the right hand side integral is very tedious. It's an example of the cyclic type so try using integration by parts twice and noticing that you end up where you started. 
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. We proceed as in the hint. Multiplying both sides by gives
Now, the left hand side is just and so
Integrating both side, we see that the integral of the right hand side above is difficult  so we compute it separately. We proceed by integration by parts. Let
and so the integral becomes
We use parts again. Let
Then, we have
This last integral on the right is the same as the integral on the left (up to a constant). Hence, we have
where . After this gigantic diversion, we return to the point. We had
and integrating both sides, we get
Dividing both sides by , we have
Now, we solve for . We plug in the initial condition and see that
and so isolating, we see that . Again integrating both sides of gives
where E is another constant. To compute E, we use the other initial condition that to see that
and so . Combining gives
and we are finished! 