Find x_0 and the largest interval, I, for which y(x) is a sol[ ]

[s3a]

"Find x_0 and the largest interval, I, for which y(x) is a sol[...]"

Homework Statement

Given that y = –2/x + x is a solution of the differential equation xy' + y = 2x, find $x_0$ and the largest interval, I, for which y(x) is a solution of the initial-value problem:

xy' + y = 2x; y($x_0$) = 1

Homework Equations

y(x) = –2/x + x
y($x_0$) = 1

The Attempt at a Solution

The largest intervals for which y(x) and y'(x) are analytic are (–∞, 0) and (0, ∞), but what is/are the largest interval for which y(x) and y'(x) are analytic for which y(x) is also a solution to the given initial–value problem?

Basically, how does forcing y(x) to be a solution to the given initial–value problem affect the interval?

Also, if one has two infinite intervals, aren't they equal (rather than one being larger)? For example, isn't (–∞, 0) just as large as (–∞, –25)? (I arbitrarily chose the number –25.)

Any help in understanding how to answer this problem correctly would be GREATLY appreciated!

 

[Simon Bridge]

To understand these questions, don't be frightened to play around a bit first.

i.e. have you checked that the y(x) given is actually a solution to the DE?
If it is then there is no question of "forcing" anything: it's a solution.

Is y(x) a unique solution? What would the general solution look like?

The initial value changes the problem, why wouldn't the interval be changed?

Note: when dealing with infinities, you want Cantor arithmetic.
i.e. while both your example intervals contain an infinite number of integers, you can equally say that one of your intervals contains 25 fewer integers than the other one. Expand that statement to the missing reals as well...

 

[HallsofIvy]

You are given that y= -2/x+ x and want $y(x_0)= 1$. So for what $x_0$ is $y(x_0)= -2/x_0+ x_0= 1$. And for purposes of this problem, interval A is "larger" than interval B if B is a proper subset of A. So (-∞, 0) is larger than (-∞, -25).

 

[s3a]

Thanks for the inputs.

I get x_0 = −1, 2.

Someone told me that the intervals for which y(x) and y'(x) are analytic for which y(x) is also a solution to the given initial-value problem

are (−∞, −1) and (2, ∞).

I understand why not when x = 0, but why is y(x) not a solution to the given initial-value problem on [−1,2], though?