Question

solve the equation dy/dy + (tan t ) y = cos t

Answer 1

\[\frac{ dy }{ dt } + (\tan t)y=\cos t\]
I have calculated an integrating factor of cost

Answer 2

...and from multiplying the original equation by cost I got
cost dy/dt + sin t y = cos^2 t

Answer 3

What did you get as your integrating factor?

Answer 4

cos(t)

Answer 5

\[\frac{ d }{ dt }(ycos(t))=\cos^2(t)\]

Answer 6

for the integrating factor...
\[e ^{\int\limits_{}^{}\frac{ \sin(t) }{ \cos(t) }dx} = e ^{\ln \cos(t)} = \cos (t) ?\]

Answer 7

yeah I got the same as you

Answer 8

sin over cos is otherwise known as tan

Answer 9

Then integrate both sides

Answer 10

But as you can't integrate cos^2(t) you need to get it in terms of cos(2t)

Answer 11

Do you know the integral of tangent x dx

Answer 12

\[\cos^2(t)=\frac{ 1 }{ 2 }(\cos(2t) +1)\]

Answer 13

cos2t = 2 cos^2t +1
so cos ^2 t = 1/2 ( cos 2t - 1)

Answer 14

* plus 1 not minus

Answer 15

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

Answer 16

Or just acknowledge tangent, simple

Answer 17

\[ycos(t)=\frac{ -1 }{ 4 }\sin(2t) +\frac{ 1 }{ 2 }t +C\]

Answer 18

\[y=\frac{ -\sin(2t) }{ 4\cos(t) }+\frac{ t }{ 2\cos(t) }+\frac{ C }{ \cos(t) }\]

Answer 19

apparently this can be simplified to y=(t+C)cos(t) according to the answer in the book??

Answer 20

Yeah, secant is your integrating factor

Answer 21

I thought it was cos(t)?

Answer 22

Oh yeah! We took cos(t) to be the intergrating factor where actually it is 1/cos(t) sorry for leading you down a rabbit hole!

Answer 23

oh, how is it 1/cos(t) ?

Answer 24

because if you integrate sin(t)/cos(t) its 1/cos(t) think about f`(x)/f(x) when you integrate it it goes to ln(1/f(x))

Answer 25

if we change the IF to \[\frac{ 1 }{ \cos(t) }\] we get that \[\frac{ 1 }{ \cos(t) }\frac{ dy }{ dy }+\frac{ \sin(t)y }{ \cos^2(t) }=1\]
\[\frac{ d }{ dt }(\frac{ y }{ \cos(t) })=1\]
\[\frac{ y }{ \cos(t) }=t+C\]