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

MATH103 April 2011

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

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[show] Hint
This differential equation is separable.

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[show] Solution
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.