Science:Math Exam Resources/Courses/MATH215/April 2014/Question 01 (b)

MATH215 April 2014

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• April 2013 • December 2011 • December 2013 •

[hide] Question 01 (b)
Solve ${\frac {dy}{dx}}=xy+x+y+1$ for $y(x)$ , subject to $y(1)=1$ .

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!

[show] Hint
Try factoring the right-hand side.

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.

If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work.
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[show] 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. Note that we can factor the right-hand side: $xy+x+y+1=(y+1)(x+1)$ . Therefore ${\frac {dy}{dx}}=(y+1)(x+1)$ and hence the problem is really just a disguised exercise in separation of variables. Continuing, we find $\ln(1+y)=\int {\frac {dy}{1+y}}=\int (x+1)dx={\frac {1}{2}}(x+1)^{2}+C_{0}$ ( $C_{0}$ a constant of integration), or solving for $y$ , $y(x)=C_{1}e^{{\frac {1}{2}}(1+x)^{2}}-1,$ (where $C_{1}=e^{C_{0}}$ ). The constant is determined from the initial condition $y(1)=1$ ; this gives $C_{1}=2e^{-2}$ . Upon simplification, we therefore conclude that $y(x)=2e^{\frac {(x+3)(x-1)}{2}}-1.$ To check our answer, we can differentiate it (exercise) to make sure it does indeed satisfy the given differential equation.