-m99.53 ty-plane hypotheses of Theorem 2.4.2.

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SUMMARY

The hypotheses of Theorem 2.4.2 are satisfied in the ty-plane except along the line defined by the equation \(2t + 5y = 0\), which corresponds to \(y = -\frac{2}{5}t\). The differential equation given is \(y' = \frac{t - y}{2t + 5y}\), and the continuity of the function \(f(t, y)\) is critical for the hypotheses to hold. The rational function is continuous wherever its denominator is non-zero, specifically where \(2t + 5y \neq 0\).

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State where in the ty-plane the hypotheses of Theorem 2.4.2 are satisfied

$\displaystyle y^\prime= \frac{t-y}{2t+5y}$
ok I don't see how this book answer was derived since not sure how to separate varibles
$2t+5y>0 \textit{ or }2t+5y<0$
 

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This is very concerning. You have posted a series of question where you seem to have no idea what the question is asking. In each one you have immediately jumped to trying to solve the differential equation when the question given does not ask you to do so.

Here the problem just asks you to "State where in the ty-plane the hypotheses of theorem 2.4.2 are satisfied". The hypotheses of theorem 2.4.2 are that the function f(t,y) in the differential equation y'= f(t,y) be continuous. Here the differential equation is $y'= \frac{t- y}{2t+ 5y}$. So the problem is asking "where is $
\frac{t- y}{2t+ 5y}$ continuous?"

You should know that a rational function, such as this, is continuous as long as the denominator is not 0. So we want to find (t, y) such that $2t+ 5y\ne 0$. The simplest way to do that is to say where it is 0! 2t+ 5y= 0 is a straight line in the ty-plane. That is equivalent to the line y= -(2/5)t, a line through the origin with slope -2/5. The hypotheses of theorem 2.4.2 are satisfied every where EXCEPT on that line.
 

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