Question

How do you determine an interval in which a solution of the given I.V.P. is certain to exist? @ganeshie8

Answer 1

(Without solving)\[y'+\frac{ y }{ \ln(t) } = \frac{ \cot(t) }{ \ln(t) }\] \[y(2) = 3\] I see that \[a(x) = \frac{ 1 }{ \ln(t) }~~~~~\ln(t) \neq 1\]
and \[b(x) = \frac{ \cot(t) }{ \ln(t) }\] this is the one I'm having a bit of trouble with, so cot is undefined at pi, and lnt at 1?

Answer 2

That should be a(t) and b(t) so \[\frac{ 1 }{ lnt } ~~~t>1\] but b(t) is where I'm having a bit of trouble

Answer 3

I know that 3 has no effect on the interval, and that y(2) we have t=2 so I guess I need to figure out where exactly b(t) is continuous

Answer 4

@amistre64 @freckles

Answer 5

\[t>0 \text{ since we have } \ln(t) \\ t \neq 1 \text{ since we have } \frac{1}{\ln(t)} \\ t \neq n \pi , n \in \mathbb{Z} \text{ since we have } \cot(t)=\frac{\cos(t)}{\sin(t)}\]

Answer 6

so we have the intervals:
(0,1)
(1,pi)
(pi,2pi)
(2pi,3pi)
,...
(npi,(n+1)pi)

but t=2 is in (1,pi)

Answer 7

Ahh, ok I see, so it's where they share an intersection, so it doesn't really matter if b(t) is set up as \[\frac{ \cot(t) }{ \ln(t) }\]

Answer 8

well we already took care of the 1/ln(t) part
so we just had to look at the cot(t) part for b(t)

Answer 9

Ok gotcha, thanks!

Answer 10

but yeah b(t)=cot(t)/ln(t) is defined and continuous on:
(0,1) U (1,pi) U (pi,2pi) U (2pi,3pi) U ....
and a(t)=1/ln(t) is defined and continuous on:
(0,1) U (1,inf)
so yes I just took the intersection

Answer 11

which was
(0,1) U (1,pi) U (pi,2pi) U (2pi,3pi) U ....

Answer 12

and from that I just looked for the interval that contained t=2

Answer 13

http://tutorial.math.lamar.edu/Classes/DE/IoV.aspx
here are some more examples

Answer 14

Thanks a lot, I think I'm starting to get it now, thanks for the link!

Answer 15

np