Science:Math Exam Resources/Courses/MATH103/April 2015/Question 05 (c)

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MATH103 April 2015

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Question 05 (c)
Consider the differential equation ${\frac {dy}{dx}}=y^{2}-1$ . (c) Solve the differential equation using the initial condition $y(0)=2$ .

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!

Hint
This equation is a separable differential equation.

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Solution
This is separable equation, and we have ${\frac {dy}{y^{2}-1}}=dx.$ So, we use the partial fraction to write ${\frac {dy}{(y+1)(y-1)}}=dx\Longrightarrow {\frac {-{\frac {1}{2}}}{y+1}}dy+{\frac {\frac {1}{2}}{y-1}}dy=dx.$ By taking integral from both sides, we get that $\int {\frac {-{\frac {1}{2}}}{y+1}}dy+\int {\frac {\frac {1}{2}}{y-1}}dy=\int 1dx\Longrightarrow -{\frac {1}{2}}\ln(y+1)+{\frac {1}{2}}\ln(y-1)=x+C_{1}.$ So, $\ln(y-1)-\ln(y+1)=2x+C_{2}\Longrightarrow \ln({\frac {y-1}{y+1}})=2x+C_{2}\Longrightarrow {\frac {y-1}{y+1}}=Ce^{2x}.$ We apply the initial condition $y(0)=2$ to obtain ${\frac {2-1}{2+1}}=Ce^{0}\Longrightarrow C={\frac {1}{3}}.$ We finally get ${\frac {y-1}{y+1}}={\frac {1}{3}}e^{2x}\Longrightarrow 3y-3=e^{2x}y+e^{2x}\Longrightarrow \color {blue}y={\frac {3+e^{2x}}{3-e^{2x}}}.$