Question

solve the initial value dy/dt=(t^2)tan(y) y(0)=0

Answer 1

this can be separated
\[\implies \frac{dy}{\tan y} = t^2 dt\]
does that help?

Answer 2

Ive already done that so the integrals are ln/sin(y)/ = (t^3)/3+C right?

Answer 3

yes

Answer 4

y(0) = 0 so change t to 0 and y to 0

Answer 5

then solve for x

Answer 6

so when you plug in y(0) = 0 the left side doesnt work

Answer 7

hmm interesting

Answer 8

because you get ln of 0 which is undefined

Answer 9

any ideas?

Answer 10

maybe c just doesnt exist?

Answer 11

okay do this
\[\sin y=e ^{t ^{3}/3}+c\]
now solve for c by putting y=0

Answer 12

so c=1?

Answer 13

sorry it will be
\[e ^{c}\]

Answer 14

\[\huge y = \sin^{-1} \left( e^{\frac{t^3} 3 + c_1} \right)\]

nothing seems to fit

Answer 15

@Igbasallote is my answer correct or will it be only c

Answer 16

should be e^c

Answer 17

but you can just call e^c as K

Answer 18

then solve for K

Answer 19

it's solveable

Answer 20

dont you get e^c=0

Answer 21

yes. e^c = 0

Answer 22

so you can just put that...

Answer 23

there's no expoennt you can raise e that will result to 0

Answer 24

so expressing it as e^c maybe the best you can hope for

Answer 25

true

Answer 26

i can just say e^c is an arbitrary constant

Answer 27

oh wait no

Answer 28

it's really impossible lol

Answer 29

the GS is \[\huge \sin y = e^{\frac{t^3}3} \times e^c\]
so if e^c is 0...this becomes 0....

Answer 30

thats what i thought

Answer 31

might be discontinuous at y(0) = 0?

Answer 32

true i didnt think of that

Answer 33

\[y(0)=0\implies y'(0)=t^2\tan(y)=(0)(0)=0\]No change in \(y\) around the point \((0,0)\), so \(y\) remains constant. Check this using a slope field:So \(y=0\) is the solution.