# Non-linear IVP question

I am knew to this site. Hopefully It will end my search.

Let us take a non-linear IVP $dy/dx = (1-2x) y^2$ with $y(0) = y_0$

Now my questions are: (A) Show that this initial value problem has a unique solution for any choice of $y_0$ and (B) How does the interval of definition of the solution depend on $y_0$ ?

Could you explain the procedure to answer those questions? please...

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For question (A), look in your notes or textbook for a general existence and uniqueness theorem. It will cover this case. Check all assumptions and apply it. For (B), solve the equation. HINT: The right hand side is already factored. –  Hans Engler Mar 6 '12 at 18:08

You can solve this by separation of variables, provided $y$ is not $0$, since you have to divide by $y$ to separate variables. If I'm not mistaken, you should get $$y = \frac{1}{x^2-x+C}.$$ The function $y=$ for all $x$, is also a solution, as you can check by substitution.
As for "intervals of definition", notice that the quadratic equation $x^2-x+C=0$ has no solutions if $C<-1/4$. If $C>-1/4$, it has two solutions, and so you get vertical asymptotes in the solution. So you might say the interval on the $x$-axis between $0$ and the nearest vertical asymptote to $0$ is the "interval of definition". But really, the two vertical asymptotes break the line into three intervals. You get solutions on each interval. But only one of the intervals contains $0$. So on that interval, the initial condition specifying $y(0)$ determines the solution; on the other two intervals it doesn't. Since $y(0)=1/C$, we get $C= 1/y(0)$.
But which of the three intervals contains $0$? That depends on $C$ and hence on $y(0)$. The solution to the quadratic equation is $$x=\frac 1 2 \pm \sqrt{\frac 1 4 + C\ {}}.$$ If $C$ is just a little bit bigger than $-1/4$, then both of these numbers are positive, so $0$ is to the left of both of them. If $C>0$ then $0$ is between the points where the vertical asymptotes are. If $C=0$ exactly, then one of the vertical asymptotes is at $0$ and there will be no solution that satisfies an initial condition saying $y(0)={}$ some finite number.