# Elementary Autonomous Initial Value Textbook Problem

• Jun 1st 2010, 07:13 PM
VinceW
Elementary Autonomous Initial Value Textbook Problem
Solve the following initial value problem and determine the interval where the solution exists.

dy/dt = 2ty^2
y(0) = y_0

When y_0 <> 0, then y = 1/(1/y_0 - t^2).
When y_0 = 0, then y = 0

When y_0 > 0, the interval where solution exists is |t| < 1/sqrt(y_0)
When y_0 <= 0, the interval where solution exists is -infinity < t < +infinity

My work:

The equation is separable:

dy/y^2 = 2t dt
-1/y = t^2 + C

When y_0 = 0, I use intuition to come to the same solution that the textbook shows.

When y_0 <> 0
C = -1/y_0
y = 1/(1/y_0 - t^2).
The interval where the solution exists is where the denominator expression does not equal zero. Algebraically, that simplifies to t <> 1/sqrt(y_0)

The textbook answer is different. What did I do wrong or omit?
• Jun 4th 2010, 05:01 AM
Diemo
So far so good, you have
$\displaystyle y=\frac{1}{\frac{1}{y_0}-t^2}$.
Now rearrange this a litlle by multiplying the top and the bottom by $\displaystyle y_0$ to get
$\displaystyle y=\frac{y_0}{1-y_0t^2}$.
Now your correct in saying that this exists provided the denominator is non-zero, or when $\displaystyle y_0t^2 \not = 1$. But note that $\displaystyle t^2$ is always greater than zero, hence when $\displaystyle y_0$ is negative, it is impossible for $\displaystyle t^2y_0$ to be positive, hence the equation exists for all $\displaystyle y_0<0, t$. For $\displaystyle y_0>0$ the denominator is equal to zero when $\displaystyle t=+-\frac{1}{\sqrt{y_0}}$.

• Jun 9th 2010, 12:47 PM
VinceW
Quote:

Originally Posted by Diemo
So far so good, you have
$\displaystyle y=\frac{1}{\frac{1}{y_0}-t^2}$.
Now rearrange this a litlle by multiplying the top and the bottom by $\displaystyle y_0$ to get
$\displaystyle y=\frac{y_0}{1-y_0t^2}$.
Now your correct in saying that this exists provided the denominator is non-zero, or when $\displaystyle y_0t^2 \not = 1$. But note that $\displaystyle t^2$ is always greater than zero, hence when $\displaystyle y_0$ is negative, it is impossible for $\displaystyle t^2y_0$ to be positive, hence the equation exists for all $\displaystyle y_0<0, t$. For $\displaystyle y_0>0$ the denominator is equal to zero when $\displaystyle t=+-\frac{1}{\sqrt{y_0}}$.

$\displaystyle y_0 > 0$
$\displaystyle t \ne \pm \frac{1}{\sqrt{y_0}}$
$\displaystyle |t| < \frac{1}{\sqrt{y_0}}$
If I set $\displaystyle y_0 = 1$ and graph the corresponding formula $\displaystyle y = \frac{1}{1-t^2}$, the graph does not exist at $\displaystyle t = \pm 1$, but exists at all other values of t.