Question

diff eq. solve equation: (siny)^2dt+tsin2ydy=0

Answer 1

looks separable to me

Answer 2

first determine if they're exact. they are.
dp/dy=2sinycosy
dq/dt=2sinycosy

Answer 3

might even be exact

Answer 4

cool lol

Answer 5

the t sin 2y might be more doable as an integration

Answer 6

sin(2y) = 2 sin(y) cos(y)

Answer 7

but then the dt part is just a sdoable lol forgot we could keep the NOTs constant

Answer 8

my final answer is not the same as the book's so idk if i'm suppose to keep going. i have F=t(siny)^2. book says lnt+2lnsiny=C. when did it become function of e??

Answer 9

\[sin^2(y)dt+t\ sin(2y)dy=0\]
\[\int sin^2(y)dt=t\ sin^2(y)+g(y)\]
\[D_y[t\ sin^2(y)+g(y)]=2t\ sin(y)cos(y)+g'(y)\]

Answer 10

since sin(2y) = 2 sin(y) cos(y) it looks like g'(y) = 0

Answer 11

yessss

Answer 12

so F=t(siny)^2y

Answer 13

g(y) = some arbitrary constant still

Answer 14

F = t sin^2(y) + C

if i read it right

Answer 15

o ok

Answer 16

idk why i have in my notes F=function=C...didn't make sense but now i know i prob copied wrong. thanks

Answer 17

youre welcome, and good luck :)

Answer 18

ok it still doesn't make sense to me. i did mean to write in my notes function=C and that's the way the book has it too. so back to the question how does it go to C=lnt+2lnsiny? no initial value given

Answer 19

well, lets see how we do when use the separable method instead of the exact method:
\[sin^2(y)\ dt+t\ sin(2y)\ dy=0\]
\[t\ sin(2y)\ dy=-sin^2(y)\ dt\]
\[\frac{sin(2y)}{-sin^2(y)}\ dy=t^{-1}\ dt\]
\[\frac{2sin(y)cos(y)}{-sin(y)sin(y)}\ dy=t^{-1}\ dt\]
\[\frac{2cos(y)}{-sin(y)}\ dy=t^{-1}\ dt\]
\[\int \left(-2\frac{cos(y)}{sin(y)}\ dy=t^{-1}\ dt\right)\]
\[-2ln(sin(y))=ln(t)+C\]

Answer 20

im sure if I divied that up the other way that Id get +s instead of -s

Answer 21

thank you amistre, i wish i can figure this out using the exact method tho. good thing about our test coming up is that the instructor will tell us which method to use for each problem...but that can be a bad thing too since we also HAVE to use the specified method

Answer 22

the separable i believe has an issue when y=0 since we get a 0 in the denominator along the way.

the exact does not have that issue:

\[sin^2(y)\ dt+t\ sin(2y)\ dy=0\]
\[\int sin^2(y)\ dt=t\ sin^2(y)+g(y)\]
\[D_y[t\ sin^2(y)+g(y)]=t\ 2sin(y)cos(y)+g'(y)\]
\[t\ 2sin(y)cos(y)=t\ sin(2y):\ t\ sin(2y)+g'(y)\]
therefore:
\[t\ sin(2y)+g'(y)=t\ sin(2y)+ 0 \]
\[\int (g'(y)=0)\ \implies \ g(y)=C \]
\[y=t\ sin(2y)+C\]
and from the separable we get:

\[-2ln(sin(y))=ln(t)+C\implies\ \frac{1}{sin^2(y)}=Ce^t\]
\[sin^2(y)=Ce^{-t} \implies \ y=sin^{-1}(Ce^{-t/2});\ y\ne 0\]

Answer 23

as long as you can show how you got to your answer; and define the interval upon which it is defined, you should be good

Answer 24

thank you thank you