Question

solve the equation dy/dt + cot(t) y = sin (t)
(Integration factor = sin x )

Answer 1

so far I have:
\[\frac{ dy }{ dx } + \cot(t)y = \sin (t)\]
cot=cos/sin
Integration factor:\[e ^{\int\limits_{}^{}\frac{ \cos(t) }{ \sin(t) }dx}\]
where
\[{\int\limits_{}^{}\frac{ \cos(t) }{ \sin(t) }dx} = \ln \sin(t)\]
therefore Integrating factor = sin(t)

Answer 2

Are you sure the integrating factor is sin(t)?
try and integrate cos(t)/sin(t) again

Answer 3

I substituted u for sin(t)
so du/dx = cos(t)
\[\int\limits_{}^{}\frac{ \cos(t) }{ \sin(t) }dt = \int\limits_{}^{} \frac{ \cos(t) }{ u }\frac{ du }{ \cos(t) } = \int\limits_{}^{}\frac{ 1 }{ u } = \ln u = \ln \sin(t)\]

Answer 4

@OllieT

Answer 5

Okay yeah that makes sense.
so multiply everything by the integrating factor

Answer 6

\[\sin(t)\frac{ dy }{ dt }+\frac{ \cos(t) }{ \sin ^{2}(t) }y=1\]

Answer 7

\[\frac{ dy }{ dx }(\sin(t)y)=1\]

Answer 8

*sorry its d/dx not dy/dx

Answer 9

are you sure what you wrote originally in the question is correct? If so you shouldn't get what you got

Answer 10

remember you multiply everything by sin(t) not divide

Answer 11

yeah, other than the fact I kept writing dx instead of dt, oops

Answer 12

so from what you wrote initially you should get \[\sin(t)\frac{ dy }{ dt }+\cos(t)y=\sin^2(t)\]

Answer 13

the answer should be y=(1/2t-1/4 sin(2t)+c)cosec(t)

Answer 14

ohhhh wait yes sorry, i made it equal to 1 instead of sin^2 x

Answer 15

so then you get \[\frac{ d }{ dt }(ysin(t))=\sin^2(t)\]

Answer 16

\[\sin ^{t}=1/2(\sin(2t)-1) ?\]

Answer 17

wait, that was completely wrong!
\[\sin ^{2}t=1/2-1/2\cos(2t)\]

Answer 18

Yes so then integrate both sides

Answer 19

Got it! finally

Answer 20

Sweet