Question

Verify by differentiation and substitution that the given function is a solution.
y' = y^2 + 4 (l t l < pi/4); y = 2 tan 2t

Answer 1

Sorry I don't know how to do this stuff ;s You can go in the math chat and ask for help.

Answer 2

The directions say to differentiate and substitute
You want to differentiate the given y = 2tan(2t) solution

Any ideas on how to take the derivative of that?

Answer 3

It's function inside of a function so what derivative rule do we need? @mony01

Answer 4

is it the chain rule?

Answer 5

@Miracrown

Answer 6

What is the derivative of the inside, 2t?

Answer 7

The easiest way to handle this problem is to first find the derivative, y', assuming the given y is a solution.
Would you like to work the derivative, y', while we observe? @mony01

Answer 8

would the derivative be 2(2tan2t)*2

Answer 9

@Miracrown

Answer 10

first find the derivative of tant then multiply this by the derivative of 2t (chain rule), then multiply the lot by 2

Answer 11

that will give you the LHS

Answer 12

@alekos the derivative of tan t is sec^2 t right?

Answer 13

yes

Answer 14

I know that since y=2tan2t I need to plug this in into the other equation right so it would be y'= (2tan2t)^2 +4 so now is it y'= (2sec^2(2t))^2 +4

Answer 15

\[y' = 2*(2\sec^2 (2t)) = 4 \sec^2 (2t)\]
\[y' = y^2 + 4\]
\[\rightarrow 4 \sec^2(2t) = (2 \tan(2t))^2 + 4\]

Answer 16

where did you put the +4 when you got 4sec^2(2t)?

Answer 17

you are confusing 2 different things

the differential equation ---> y' = y^2 + 4

and y' as simply the derivative of y, which is given as 2tan(2t)

they are different but equal, that is why we can substitute it into the diff equation

Answer 18

oh so the whole point is to solve for y' using y only

Answer 19

yes ... they gave you what the solution "y" is
to verify you have to plug it into the diff equation but you also have to plug in something for y'
thus taking derivative of y gives y'

no solving at all really

Answer 20

oh thank you i understand now so we didn't even have to deal with the +4 right

Answer 21

well its still part of the diff equation but it has nothing to do with y'

Answer 22

ok thank you so much

Answer 23

yw :)