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anonymous
 4 years ago
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?
anonymous
 4 years ago
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?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i am confused because integral of 1/(1+t^2) is arctant...

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1if y=tan(t) IS a solution then itll fit in just nicely

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok, cuz it actually if a solution...but shouldn't the question have assigned different variables??

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1no, t is t is t in all the derivatives of y

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well 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
 4 years ago
Best ResponseYou've already chosen the best response.1well, tan(t) is that has a zero in it, and is contiuous would be: (pi/2 , pi/2) if i see this right

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1you may be reading to much into the problem :)

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0though the interval pi/2,pi/2 still makes sense

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i 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
 4 years ago
Best ResponseYou've already chosen the best response.0arctan(t) is undefined at \[\pi/2+n\pi\]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1is y=tan(t) a solution?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1or is it; y=arctan(t) a solution? are you reading it correctly is the quandry

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=%20y'%3D1%2F(1%2Bt%5E2)&t=crmtb01 in case you didn't know the derivative of arctan...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i 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
 4 years ago
Best ResponseYou've already chosen the best response.1as is, we are stating: \[tan'(t)=\frac{1}{1+t^2}\] \[sec^2(t)=\frac{1}{1+t^2}\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0no I said earlier it should be arctan tan is not a solution

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1if not a solution; then there is no interval to worry about since no interval is a solution to a nonsolution :)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1\[\frac{dy}{dt}=\frac{1}{1+t^2}\] \[{dy}=\frac{1}{1+t^2}dt\] \[\int\ ({dy}=\frac{1}{1+t^2}dt)\] \[y=tan^{1}(t)+C_1\]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1when y=0 and t=0; we get: \[0=tan^{1}(0)+C_1\] \[0=C_1\] therefore, given the IVP \[y=tan^{1}(t)\]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1this shows that y=tan(t) is NOT a solution to the IVP and is therefore no need to worry about any nonexisting intervals
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