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#### alane1994

##### Active member

- Oct 16, 2012

- 126

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!