# Math Help - Elementary Autonomous Initial Value Textbook Problem

1. ## 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?

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

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

When
$y_0 > 0$

then your work and my work indicate that the solution exists for
$t \ne \pm \frac{1}{\sqrt{y_0}}$

the book says that the solution exists for
$|t| < \frac{1}{\sqrt{y_0}}$

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

It would seem that the book is simply wrong? For one question I wouldn't be surprised (I find at least one or two mistakes in every text book), but there were three questions in a row that were differing from my answers like this and that usually means I am making the mistake rather than the book.