Science:Math Exam Resources/Courses/MATH103/April 2011/Question 06 (b)

MATH103 April 2011

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Question 06 (b)
Consider the differential equation ${\frac {dy}{dx}}=(y^{2}-1)x.$ Solve for $y(t)$ given the initial value $y(0)=0$.

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 differential equation is separable.

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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. The differential equation is separable; that is, ${\begin{aligned}{\frac {dy}{dx}}=(y^{2}-1)x\implies \int {\frac {1}{y^{2}-1}}\,dy=\int x\,dx.\end{aligned}}$ On the right hand side, ${\begin{aligned}\int x\,dx={\frac {x^{2}}{2}}+C_{1},\end{aligned}}$ for some constant $C_{1}.$ On the left hand side, we factor $y^{2}-1=(y+1)(y-1)$ and use partial fractions. ${\begin{aligned}&{\frac {1}{(y+1)(y-1)}}={\frac {A}{y+1}}+{\frac {B}{y-1}}\\&\implies 1=A(y-1)+B(y+1).\end{aligned}}$ This must be true for all $y$. In particular, at $y=-1$ , $1=-2A$, hence, $A=-1/2$. At $y=1$, $1=2B$, hence $B=1/2$. Therefore, ${\begin{aligned}\int {\frac {1}{y^{2}-1}}\,dy&=-{\frac {1}{2}}\int {\frac {1}{y+1}}\,dy+{\frac {1}{2}}\int {\frac {1}{y-1}}\,dy\\&=-{\frac {1}{2}}\ln |y+1|+{\frac {1}{2}}\ln |y-1|+C_{2}\\&={\frac {1}{2}}\ln \left|{\frac {y-1}{y+1}}\right|+C_{2},\end{aligned}}$ for some constant $C_{2}.$ The initial condition is $y(0)=0,$ so at least for small $x$ , we know that $-1<y(x)<1$. This means that y-1 is negative and y+1 is positive. Hence $\left|{\frac {y-1}{y+1}}\right|=-{\frac {y-1}{y+1}}={\frac {1-y}{y+1}}={\frac {1-y}{1+y}}.$ Therefore ${\begin{aligned}\int {\frac {1}{y^{2}-1}}\,dy&={\frac {1}{2}}\ln \left({\frac {1-y}{1+y}}\right)+C_{2}.\end{aligned}}$ Equating the left and right sides of our first equation, we have ${\begin{aligned}&{\frac {1}{2}}\ln \left({\frac {1-y}{1+y}}\right)={\frac {x^{2}}{2}}+C_{1}-C_{2}\\&\implies {\frac {1-y}{1+y}}=Ae^{x^{2}}\\&\implies 1-y=(1+y)Ae^{x^{2}}\\&\implies 1-Ae^{x^{2}}=y(1+Ae^{x^{2}})\\&\implies y={\frac {1-Ae^{x^{2}}}{1+Ae^{x^{2}}}}.\end{aligned}}$ where $A=e^{2(C_{1}-C_{2})}$. From the initial condition, $y(0)=0$, $A$ must be equal to 1. Hence, ${\begin{aligned}y(x)={\frac {1-e^{x^{2}}}{1+e^{x^{2}}}}.\end{aligned}}$ Note that this solution does satisfy $-1<y(x)<1$ for all $x$. Remember: It is important (and usually easy) to check that your solution actually satisfies the differential equation.

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