Thursday

January 29, 2015

January 29, 2015

Posted by **Erica** on Thursday, January 31, 2013 at 3:41pm.

a) Show that the constant function y1(t)=0 is a solution.

b)Show that there are infinitely many other functions that satisfy the differential equation, that agree with this solution when t<=0, but that are nonzero when t>0 [Hint: you need to define these functions using language like " y(t)=...when t<=0 and y(t)=...when t>0 and "]

c) Why doesn't this example contradict the Uniqueness Theorem?

I'm trying to do part b and after I separated and integrated I got

ln|y|=(-1/t)+C

I'm not sure if I can get C with the solution they gave in part a)y1(t)=0.

Anyways, I get y(t)=Ce^-(1/t). I don't know where to go from there.

- Differential Equations -
**Damon**, Thursday, January 31, 2013 at 3:52pmdy/y = dt/t^2

ln y = -1/t + c

y = e^(-1/t+c) = e^c e^(-1/t)

=C e^(-1/t) agree

C can be anything so

what if C = 0 ?

then y(t) = 0 for all t

if C = 0 for t</= 0 then C can be anything at all for t>0

- Differential Equations -
**Erica**, Thursday, January 31, 2013 at 3:58pmIf y(t) depends on what C is, then how this equation doesn't contradict the uniqueness theorem if it has many solutions?

- Differential Equations -
**Damon**, Thursday, January 31, 2013 at 4:20pmBecause the general solution contains an an arbitrary constant C. The value of C depends on your boundary conditions, for example if y =5 at t = 2

then

5 = C e^-(1/2)

5 = C (.606)

C = 8.24

NOW your solution is unique.

- Differential Equations -
**Damon**, Thursday, January 31, 2013 at 4:34pmscroll down through this:

http://hyperphysics.phy-astr.gsu.edu/hbase/diff.html

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