- #1
dorado29
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I'm working on a problem for my diffEQ's class. It reads:
For each of the given initial value problems, what is the region where the existence and uniqueness of the solution is guaranteed? Is it all t, or t > -3, or a neighborhood of t = 1?
a) x'' = t x x' + sin(t)
b) x'' = tx + 2x' + sin(t)
c) (t+3)x'' = tx +2x' + sin(t)
The initial values given for a, b and c are all x(1) = 5 and x'(1) = 2
My thoughts/ attempts:
It seems that a, b and c are all linear. In my book (my professor wrote it), it says that
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
If f(t, x, x') is a continuously differentiable function, then the initial value problem has exactly one solution for t close to t0. Furthermore, this solution exists for all t, if x'' = f(t, x, x') is a linear function.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
I don't see any x or x' in a, b, or c that aren't to the first power (I believe that would make them all linear?). The way I'm interpreting the above bit from the book, it would seem that the existence is guaranteed for all t values. I don't think this is right because when equation c is reordered to get x'' on the left and everything else to the right, there is a (t + 3) term in the denominator. Would this mean it's guaranteed for any value of t other than -3 to avoid a 0 denominator value?
I'm unsure if multiplying x and x' together like in equation a would make it nonlinear..
Also, what do you guys think "or a neighborhood of t = 1" means? The word neighborhood throws me off..
For each of the given initial value problems, what is the region where the existence and uniqueness of the solution is guaranteed? Is it all t, or t > -3, or a neighborhood of t = 1?
a) x'' = t x x' + sin(t)
b) x'' = tx + 2x' + sin(t)
c) (t+3)x'' = tx +2x' + sin(t)
The initial values given for a, b and c are all x(1) = 5 and x'(1) = 2
My thoughts/ attempts:
It seems that a, b and c are all linear. In my book (my professor wrote it), it says that
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
If f(t, x, x') is a continuously differentiable function, then the initial value problem has exactly one solution for t close to t0. Furthermore, this solution exists for all t, if x'' = f(t, x, x') is a linear function.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
I don't see any x or x' in a, b, or c that aren't to the first power (I believe that would make them all linear?). The way I'm interpreting the above bit from the book, it would seem that the existence is guaranteed for all t values. I don't think this is right because when equation c is reordered to get x'' on the left and everything else to the right, there is a (t + 3) term in the denominator. Would this mean it's guaranteed for any value of t other than -3 to avoid a 0 denominator value?
I'm unsure if multiplying x and x' together like in equation a would make it nonlinear..
Also, what do you guys think "or a neighborhood of t = 1" means? The word neighborhood throws me off..