Question

Find the general solution.

Answer 1

\[\frac{ dr }{ d \theta } + r*\sec(\theta) = \cos(\theta)\]

Answer 2

Hint: \[\int\limits \sec(\theta)*d \theta = \int\limits \sec(\theta) * \frac{ \sec(\theta) + \tan(\theta) }{ \sec(\theta) + \tan(\theta) } *d \theta\]

Answer 3

lol

Answer 4

For anyone bored of the usual questions.

Answer 5

well your hint was to the integrating factor which is sec(theta)+tan(theta)
and this is a first order linear differentiation equation

Answer 6

Hint 2: Use integrating factor,

\[u(\theta) = e ^{\int\limits \sec( \theta) d \theta}\]

Answer 7

yea

Answer 8

call it v (integrating factor that is)
(rv)'=vcos(theta)
integrate both sides

Answer 9

yes

Answer 10

Most of the work on this was just showing the integrations, not from the tables of integrals.

Answer 11

yea no tables needed

Answer 12

\[r(\theta) = \dfrac{\int \cos \theta e^{\int \sec \theta d\theta}~d\theta +C}{e^{\int \sec \theta d\theta}}\]

Answer 13

because 1+sin(theta) is pretty elementary

Answer 14

integrating that is

Answer 15

I mean for int sec(theta) = ln(sec(x) + tan(x))

Answer 16

That one got me back in the day, without knowing to multiply by (sec + tan)/(sec + tan)

Answer 17

you can integrate csc(x) in a similar way

Answer 18

but i think you already know that
and i think you are giving us fun questions right?

Answer 19

thats a clever trick which i wouldnt have thought of w/o somebody telling me it has to be done like that !

Answer 20

yeah, just bored of the usual, find the equation for a line.

Answer 21

where are all the calculus questions :(

Answer 22

Final answer is:

\[r(\theta) = \frac{ \theta - \cos(\theta) }{ \sec(\theta) +\tan(\theta) } + \frac{ C }{ \sec(\theta) + \tan(\theta) }\]

Answer 23

Like you said:

\[[(\sec \theta + \tan \theta ) * r] ' = [\sec(\theta) + \tan(\theta) ]*\cos(\theta) = 1 + \sin \theta\]

Answer 24

\[(\sec \theta + \tan \theta)*r = \int\limits d \theta + \int\limits \sin \theta d \theta\]
\[r(\theta)(\sec \theta + \tan \theta) = \theta - \cos(\theta) + C\]

Answer 25

it would be interesting to think of a physical situation that this DE models

Answer 26

yea, it was just an old homework prob to practice the method. It probably has to do with something, not sure tough.

Answer 27

We did radioactive decay, logistic growth, conduction/diffusion models, and mixtures, as a few applications of Linear first order

Answer 28

Of course the Mass-Sprin-Dashpot system for second order

Answer 29

spring*

Answer 30

Orthogonal Trajectories.

Answer 31

\[\frac{ dr }{ d \theta } + r*\sec(\theta) = \cos(\theta)\]

\[\frac{ dr }{ d \theta } = - r*\sec(\theta) + \cos(\theta)\]

i guess it has to do with radial distance of a moving particle with respect to angle/time or something.. bit hard to visualize :O

Answer 32

salt brine solution problems fall under conduction/diffusion section is it?

Answer 33

Honestly don't remember doing too many word problems involving differential equations.
If I ever scare my ADD away for like a day I might post some so you can help me @DanJS ,

Answer 34

Mixtures type problem: like if you have a tank with a certain concentration, and more is being pumped in with a different concentration. What is the amount of salt in the tank at time t.

Answer 35

d(salt)/dt = (rate in)(concentration in) - (rate out)(concentration out)

Answer 36

If i remember right, been a couple years

Answer 37

Oh yes these are mixtures

Answer 38

Yeah, ask whenever @freckles , i am always good for a nice review.

Answer 39

Orthogonal Trajectories.