Differential Equations: Largest rectangle on ty-plane

[Northbysouth]

Homework Statement

Find the largest open rectangle in the ty-plane that contains the initial value point and satisfies the following theorem:

Let R be the open rectangle defined by a<t<b, α<y<β. Let f(t,y) be a function of two variables defined on R where f(t,y) and the partial derivative ∂f/∂y are continuous on R. Suppose (t₀, y₀) is a point in R. Then there is an open interval t-interval (c,d) contained in (a, b) and containing t₀, in which there exists a unique solution of the initial value problem

y'f(t,y), y(t₀) = y₀


(t-a)(t-3)y'= 1+ sec(y) and y(π)=1

Homework Equations

The Attempt at a Solution

So, when I take the partial derivative I get

∂f(t,y)/∂y = [sec(y)tan(y)](t-1)(t-3)/(t-1)(t-3)²

From this I can see that:

t≠1
t≠3

But y is available for all values.

Hence the largest rectangle goes from 1 to 3 on the t axis and from -∞ to +∞ on the y axis. But this doesn't contain the initial value point. Am i missing something?

 

[mfb]

But this doesn't contain the initial value point.

In that case, you need a different rectangle.

 

[haruspex]

y'f(t,y), y(t₀) = y₀

I assume that should read y'=f(t,y)

(t-a)(t-3)y'= 1+ sec(y) and y(π)=1

Later you have t-1, not t-a. Which is it?

∂f(t,y)/∂y = [sec(y)tan(y)](t-1)(t-3)/(t-1)(t-3)²

Why not just sec(y)tan(y)/((t-1)(t-3))?

From this I can see that:
t≠1
t≠3
Hence the largest rectangle goes from 1 to 3 on the t axis

Think outside the box