Astrophysics
  • Astrophysics
How do you determine an interval in which a solution of the given I.V.P. is certain to exist? @ganeshie8
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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Astrophysics
  • Astrophysics
(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?
Astrophysics
  • Astrophysics
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
Astrophysics
  • Astrophysics
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

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Astrophysics
  • Astrophysics
@amistre64 @freckles
freckles
  • freckles
\[t>0 \text{ since we have } \ln(t) \\ t \neq 1 \text{ since we have } \frac{1}{\ln(t)} \\ t \neq n \pi , n \in \mathbb{Z} \text{ since we have } \cot(t)=\frac{\cos(t)}{\sin(t)}\]
freckles
  • freckles
so we have the intervals: (0,1) (1,pi) (pi,2pi) (2pi,3pi) ,... (npi,(n+1)pi) but t=2 is in (1,pi)
Astrophysics
  • Astrophysics
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 \[\frac{ \cot(t) }{ \ln(t) }\]
freckles
  • freckles
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)
Astrophysics
  • Astrophysics
Ok gotcha, thanks!
freckles
  • freckles
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
freckles
  • freckles
which was (0,1) U (1,pi) U (pi,2pi) U (2pi,3pi) U ....
freckles
  • freckles
and from that I just looked for the interval that contained t=2
freckles
  • freckles
http://tutorial.math.lamar.edu/Classes/DE/IoV.aspx here are some more examples
Astrophysics
  • Astrophysics
Thanks a lot, I think I'm starting to get it now, thanks for the link!
freckles
  • freckles
np

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