anonymous
  • anonymous
show that y=tant satisfies the IVP y'=1/(1+t^2), y(0)=0. what is the largest open interval containing t=0 over which y=tant is a solution?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
i am confused because integral of 1/(1+t^2) is arctant...
amistre64
  • amistre64
if y=tan(t) IS a solution then itll fit in just nicely
anonymous
  • anonymous
ok, cuz it actually if a solution...but shouldn't the question have assigned different variables??

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anonymous
  • anonymous
*is a solution
amistre64
  • amistre64
no, t is t is t in all the derivatives of y
anonymous
  • anonymous
well how do i find the limit of t now? both y and y' have to considered right? so i can find the largest interval?
amistre64
  • amistre64
well, tan(t) is that has a zero in it, and is contiuous would be: (-pi/2 , pi/2) if i see this right
anonymous
  • anonymous
right
amistre64
  • amistre64
you may be reading to much into the problem :)
TuringTest
  • TuringTest
but y=tan^(-1)t...
TuringTest
  • TuringTest
though the interval -pi/2,pi/2 still makes sense
anonymous
  • anonymous
i think so too cuzi still don't understand... so limit of t in y' would be neg inf to pos inf but y is restricted so in the end, largest interval is -pi/2 to pi/2?
TuringTest
  • TuringTest
arctan(t) is undefined at \[\pi/2+n\pi\]
amistre64
  • amistre64
is y=tan(t) a solution?
TuringTest
  • TuringTest
no, y'=sec^2t
anonymous
  • anonymous
yes
TuringTest
  • TuringTest
?
anonymous
  • anonymous
?
anonymous
  • anonymous
ok, no
amistre64
  • amistre64
or is it; y=arctan(t) a solution? are you reading it correctly is the quandry
TuringTest
  • TuringTest
http://www.wolframalpha.com/input/?i=%20y'%3D1%2F(1%2Bt%5E2)&t=crmtb01 in case you didn't know the derivative of arctan...
anonymous
  • anonymous
i do know the derivative that's why i was getting confused. cuz the y in the prolem doesn't equal integral of y'. ok so no, not a solution...so limit IS neg inf to pos inf
amistre64
  • amistre64
as is, we are stating: \[tan'(t)=\frac{1}{1+t^2}\] \[sec^2(t)=\frac{1}{1+t^2}\]
TuringTest
  • TuringTest
no I said earlier it should be arctan tan is not a solution
amistre64
  • amistre64
if not a solution; then there is no interval to worry about since no interval is a solution to a nonsolution :)
anonymous
  • anonymous
right
TuringTest
  • TuringTest
right!
anonymous
  • anonymous
but arctan(0)=0
amistre64
  • amistre64
\[\frac{dy}{dt}=\frac{1}{1+t^2}\] \[{dy}=\frac{1}{1+t^2}dt\] \[\int\ ({dy}=\frac{1}{1+t^2}dt)\] \[y=tan^{-1}(t)+C_1\]
anonymous
  • anonymous
mmhmm
anonymous
  • anonymous
C=0 so y=arctant
amistre64
  • amistre64
when y=0 and t=0; we get: \[0=tan^{-1}(0)+C_1\] \[0=C_1\] therefore, given the IVP \[y=tan^{-1}(t)\]
anonymous
  • anonymous
ok
amistre64
  • amistre64
this shows that y=tan(t) is NOT a solution to the IVP and is therefore no need to worry about any nonexisting intervals
anonymous
  • anonymous
ok thank you so much

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