MATH103 April 2011
• Q1 (a) • Q1 (b) • Q1 (c) • Q2 (a) • Q2 (b) • Q2 (c) • Q2 (d) • Q3 (a) • Q3 (b) • Q4 (a) • Q4 (b) • Q5 (a) • Q5 (b) • Q5 (c) • Q5 (d) • Q6 (a) • Q6 (b) • Q7 (a) • Q7 (b) • Q7 (c) • Q8 •
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.

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.
 If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

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 $y^{2}1=(y+1)(y1)$ and use partial fractions.
 ${\begin{aligned}&{\frac {1}{(y+1)(y1)}}={\frac {A}{y+1}}+{\frac {B}{y1}}\\&\implies 1=A(y1)+B(y+1).\end{aligned}}$
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,
 ${\begin{aligned}\int {\frac {1}{y^{2}1}}\,dy&={\frac {1}{2}}\int {\frac {1}{y+1}}\,dy+{\frac {1}{2}}\int {\frac {1}{y1}}\,dy\\&={\frac {1}{2}}\ln y+1+{\frac {1}{2}}\ln y1+C_{2}\\&={\frac {1}{2}}\ln \left{\frac {y1}{y+1}}\right+C_{2},\end{aligned}}$
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 y1 is negative and y+1 is positive. Hence
 $\left{\frac {y1}{y+1}}\right={\frac {y1}{y+1}}={\frac {1y}{y+1}}={\frac {1y}{1+y}}.$
Therefore
 ${\begin{aligned}\int {\frac {1}{y^{2}1}}\,dy&={\frac {1}{2}}\ln \left({\frac {1y}{1+y}}\right)+C_{2}.\end{aligned}}$
Equating the left and right sides of our first equation, we have
 ${\begin{aligned}&{\frac {1}{2}}\ln \left({\frac {1y}{1+y}}\right)={\frac {x^{2}}{2}}+C_{1}C_{2}\\&\implies {\frac {1y}{1+y}}=Ae^{x^{2}}\\&\implies 1y=(1+y)Ae^{x^{2}}\\&\implies 1Ae^{x^{2}}=y(1+Ae^{x^{2}})\\&\implies y={\frac {1Ae^{x^{2}}}{1+Ae^{x^{2}}}}.\end{aligned}}$
where $A=e^{2(C_{1}C_{2})}$. From the initial condition, $y(0)=0$, $A$ must be equal to 1. Hence,
 ${\begin{aligned}y(x)={\frac {1e^{x^{2}}}{1+e^{x^{2}}}}.\end{aligned}}$
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.

Click here for similar questions
MER QGH flag, MER QGQ flag, MER QGS flag, MER RT flag, MER Tag Initial value problem, MER Tag Separation of variables