Elementary Autonomous Initial Value Textbook Problem

Jun 2010
59
4
Solve the following initial value problem and determine the interval where the solution exists.

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

Textbook answer is:

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?
 
Apr 2009
73
12
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 \(\displaystyle t=+-\frac{1}{\sqrt{y_0}}\).

That agree with your book?
 
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Jun 2010
59
4
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 \(\displaystyle t=+-\frac{1}{\sqrt{y_0}}\).

That agree with your book?
When
\(\displaystyle y_0 > 0\)

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

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

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.

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.