# How could I find the interval where the solution is valide for the initial value problem : $y'=\frac{1+3x^2}{3y^2-6y}$ with $y(0)=1$?

How could I find the interval where the solution is valide for the initial value problem : $y'=\frac{1+3x^2}{3y^2-6y}$ with $y(0)=1$?

I think I have to consider the points where the curve-solution admit a vertical tangent.

I got $y^3-3y^2-x-x^3+2=0$ for the curve solution. According to the solution sheet, this interval should be $|x|<1$. Any help?

EDIT: I found a solution on https://www.math.psu.edu/shen_w/250/Notes/only_ch2.pdf. when we put it into google. However I don't understand why we work on interval instead of discrete points.

Thanks!

• The differential equation is only valid on the following continuous intervals: 1) $-\infty < y < 0$, 2) $0 < y < 2$, and 3) $2 < y < \infty$. It might be valid on more but since $1 + 3x^2 = 0$ permits no real solutions, there's no need to worry about possible removeable discontinuities. – Jared Feb 24 '16 at 3:35
• Why do we work on interval instead of discret points, because $3y(y-2)=0$ when $y=0, 2$? – user314567 Feb 24 '16 at 3:41

See my comment:

The differential equation is only valid on the following continuous intervals: 1) $-\infty < y < 0$, 2) $0 < y < 2$, and 3) $2 < y < \infty$. It might be valid on more but since $1 + 3x^2 = 0$ permits no real solutions, there's no need to worry about possible removeable discontinuities.

Find the "solution" first (which you did):

\begin{align} (3y^2 - 6y)dy =&\ (1 + 3x^2)dx \\ y^3 - 3y^2 =&\ x + x^3 + C \end{align}

Plug in $y = 1$ and $x = 0$ to find:

$$1 - 3 = C \rightarrow C = -2$$

This gives the following:

$$y^3 - 3y^2 = x + x^3 - 2$$

(this is what you found)

Since $y(0) = 1$ we know that we must be on the interval $0 < y < 2$. Find the min and max on this interval: max of $f(y) = y^3 - 3y^2$ on the interval $0 < y < 2$. We already know that the critical points will be $y = 0$ and $y = 2$--the end points. So just input those two numbers. This tells us that the right side ranges from $y = 0$, $0^3 - 3\cdot0^3 = 0$ to $y = 2$, $2^3 - 3\cdot2^2 = 8 - 12 = -4$. So the $x$ values, $x + x^3 - 2$, range from $-4$ to $0$.

We need to find the continuous interval of $f(x) = x + x^3 - 2$ which is bounded by $-4 < f(x) < 0$. You can verify that there are no (real) extrema (meaning it's either monotonically increasing or decreasing) of the function $f(x) = x + x^3 - 2$ by showing that $f'(x) = 1 + 3x^2 = 0$ has no real solutions. This means that the interval is simply finding the two following solutions:

$$x + x^3 - 2 = -4 \\ x + x^3 - 2 = 0$$

Clearly $x = 1$ solves $x + x^3 - 2 = 0$. Also it's clear that $x = -1$ solves $x + x^3 - 2 = -4$. This means the $x$ interval is $-1 < x < 1$ or $|x| <1$. There is no "algebraic" way (that I'm aware of) to solve those two equations other than Rational Zeros Theorem (which means your particular problem is very special--i.e. it admits rational solutions, which will not generally be the case). Otherwise, it requires numerical methods.

• And actually that's not true, what I said about no algebraic way (and I actually was vaguely aware of it). There is a "cubic formula" just as there is a "quadratic formula" although (if you look) the cubic formula is quite complicated (and I've never used it). – Jared Feb 25 '16 at 1:14

$\frac{dy}{dx}$ is undefined at $y= 0$ and $y=2$. Your initial point is $y(0)=1$.

Think about it this way: you start from the point $(0,1)$ and manually integrate to the right from there (like Euler's method). When you do that, you will realize that you are forced to stop at $(1,0)$, the slope there is undefined. The "solution" beyond that point would not really be a correct solution because it does not show this.

Same the other way. You start at $(0,1)$ and draw the graph to the left. You are forced to stop at $(-1,2)$ for the same reason.

So, the solution ($y^3 - 3y^2 -x -x^3 +2 =0$) is valid only on the interval $(-1,1)$, or $|x|<1$