Where Do Theorem 2.4.2 Conditions Apply for \(y'=\frac{t-y}{2t+5y}\)?

[alane1994]

Greetings all,

I have a diff.eq. question. I am going to put the question, the related theorem from my book, and an example from the text that I think applies.

In each of problems 7 through 12, state where in the ty-plane the hypotheses of Theorem 2.4.2 are satisfied.
ACTUAL QUESTION:
\[y\prime=\dfrac{t-y}{2t+5y}\]THEOREM 2.4.2
Let the functions \(f\) and \(\dfrac{\partial f}{\partial y}\) be continuous in some rectangle \(\alpha < t < \beta \), \(\gamma < y < \delta\) containing the point \((t_0,y_0)\). Then, in some interval \(t_0-h < t < t_0+h\) contained in \(\alpha < t < \beta\), there is a unique solution \(y=\Phi(t)\) of the initial value problem
\[y\prime=f(t,y),~~~~y(t_0)=y_0\]

EXAMPLE:
Solve the initial value problem
\[y\prime=y^2,~~~y(0)=1\]
and determine the interval in which the solution exists.
Theorem2.4.2 guarantees that this problem has a unique solution since \(f(t,y)=y^2\) and \(\dfrac{\partial f}{\partial y}=2y\) are continuous everywhere. To find the solution, we separate the variables and integrate with the result of that
\[t^{-2}dy=dt\]
and
\[-y^{-1}=t+c\]
Then, solving for y, we have
\[y=-\dfrac{1}{1-t}\].
To satisfy the initial condition, we must chose \(c=-1\), so
\[y=\dfrac{1}{1-t}\]
is the solution of the given initial value problem. Clearly, the solution become unbounded as \(t\rightarrow 1\); therefore the solution...
The rest of the example is available, however I do not believe it really applies to this problem. And this is the best looking example as well.

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Perhaps it is all the technical "stuff" that is confusing me, but I am lost for where to start.
Any and all guidance would be appreciated! :D

 

[MarkFL]

The example solution seems to have a couple of typos. If I were going to solve the given IVP, I would write:

\(\displaystyle y'=y^2\) where \(\displaystyle y(0)=1\)

Separate variables, noting that the trivial solution \(\displaystyle y\equiv0\) is lost:

\(\displaystyle y^{-2}\,dy=dt\)

\(\displaystyle -y^{-1}=t+c\)

\(\displaystyle y=\frac{1}{c-t}\)

Using the initial values, we find $c=1$, hence:

\(\displaystyle y=\frac{1}{1-t}\)

And yes, as $t\to1$ $y$ becomes unbounded.

As for the actual question you are trying to solve...I have to run now. :D

 

[Opalg]

I have a diff.eq. question. I am going to put the question, the related theorem from my book, and an example from the text that I think applies.

In each of problems 7 through 12, state where in the ty-plane the hypotheses of Theorem 2.4.2 are satisfied.
ACTUAL QUESTION:
\[y\prime=\dfrac{t-y}{2t+5y}\]THEOREM 2.4.2
Let the functions \(f\) and \(\dfrac{\partial f}{\partial y}\) be continuous in some rectangle \(\alpha < t < \beta \), \(\gamma < y < \delta\) containing the point \((t_0,y_0)\). Then, in some interval \(t_0-h < t < t_0+h\) contained in \(\alpha < t < \beta\), there is a unique solution \(y=\Phi(t)\) of the initial value problem
\[y\prime=f(t,y),~~~~y(t_0)=y_0\]

Please look carefully at the ACTUAL QUESTION (especially that sentence in blue)! You are not being asked to solve this differential equation. All that you have to do is to specify some rectangle in which the function $f(t,y) = \dfrac{t-y}{2t+5y}$ and its partial derivative $\dfrac{\partial f}{\partial y}$ are continuous. In other words, given some initial point $(t_0,y_0)$ (which the question does not specify) you need to find a rectangle that contains it and does not meet the line $2t+5y=0$ where $f$ and its derivatives have discontinuities.

If you want to go on and solve the equation, then there are techniques for doing so, but I am sure that you are not expected to do that.