Question

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?

Answer 1

i am confused because integral of 1/(1+t^2) is arctant...

Answer 2

if y=tan(t) IS a solution then itll fit in just nicely

Answer 3

ok, cuz it actually if a solution...but shouldn't the question have assigned different variables??

Answer 4

*is a solution

Answer 5

no, t is t is t in all the derivatives of y

Answer 6

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?

Answer 7

well, tan(t) is that has a zero in it, and is contiuous would be:

(-pi/2 , pi/2) if i see this right

Answer 8

right

Answer 9

you may be reading to much into the problem :)

Answer 10

but y=tan^(-1)t...

Answer 11

though the interval -pi/2,pi/2 still makes sense

Answer 12

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?

Answer 13

arctan(t) is undefined at \[\pi/2+n\pi\]

Answer 14

is y=tan(t) a solution?

Answer 15

no, y'=sec^2t

Answer 16

yes

Answer 17

?

Answer 18

?

Answer 19

ok, no

Answer 20

or is it; y=arctan(t) a solution?

are you reading it correctly is the quandry

Answer 21

http://www.wolframalpha.com/input/?i=%20y'%3D1%2F(1%2Bt%5E2)&t=crmtb01
in case you didn't know the derivative of arctan...

Answer 22

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

Answer 23

as is, we are stating:
\[tan'(t)=\frac{1}{1+t^2}\]
\[sec^2(t)=\frac{1}{1+t^2}\]

Answer 24

no I said earlier it should be arctan
tan is not a solution

Answer 25

if not a solution; then there is no interval to worry about since no interval is a solution to a nonsolution :)

Answer 26

right

Answer 27

right!

Answer 28

but arctan(0)=0

Answer 29

\[\frac{dy}{dt}=\frac{1}{1+t^2}\]
\[{dy}=\frac{1}{1+t^2}dt\]
\[\int\ ({dy}=\frac{1}{1+t^2}dt)\]
\[y=tan^{-1}(t)+C_1\]

Answer 30

mmhmm

Answer 31

C=0 so y=arctant

Answer 32

when y=0 and t=0; we get:
\[0=tan^{-1}(0)+C_1\]
\[0=C_1\]
therefore, given the IVP
\[y=tan^{-1}(t)\]

Answer 33

ok

Answer 34

this shows that y=tan(t) is NOT a solution to the IVP and is therefore no need to worry about any nonexisting intervals

Answer 35

ok thank you so much