Question

Solve the 2nd order non-homogenous ODE Initial value problem:
y''(t)+y(t) = sec(t)
y(0)=1 , y'(0)=2

Answer 1

Anyone familiar with Diff Eq?

Answer 2

This is slightly tricky question. Have your learnt yet about turning 2nd order equations into a system of first order equations? If not, what other methods have you learnt?

Answer 3

im up to Laplace transform however this is in the chapter review for undefined coefficients.

Answer 4

Ok. So use that method. The homogeneous solutions of course are sin x and cos x, hence the particular solution is of the form

yp = u(x) sin x + v(x) cos x.

Now grind away ...

Answer 5

yeah i got the homogeneous sol'n however i dont know where to start after that... nothing in this chapter deals with sec,csc,tan,cot.... so do i just treat it as 1/cos?

Answer 6

the homogenous sol'n i got was:
\[c _{1}e ^{0t}\cos(-t)+c _{2}e ^{0t}\sin(-t)\]

Answer 7

y''+y = 0 ; roots are complex; r(1+i1, 1-i1)
\[y=e^t(c_1cos(t)+c_2sin(t))\] right?

Answer 8

No. y = c1.cos t + c2.sin t

Answer 9

\[r(a+ib,a-ib)\]
\[y=e^{ax}(c_1cos(bx)+c_2sin(bx))\]
is what i recall

Answer 10

amistre those roots are incorrect. using quad. equations since 'b' = 0 there is no real part asociated

Answer 11

hmmm ..... im open to being wrong :)

Answer 12

Amistre, for heaven's sake, with y = e^(rt), the characteristic equation is r^2 + 1 = 0 and therefore r = +-i.

Answer 13

oh ;)

Answer 14

but the main problem is i dont know what to 'guess' as the y_p

Answer 15

\[r^2+1=(r+i)(r-i)\]
i had the right thought, my fingers just betrayed me lol

Answer 16

would it be Asec(t)+Bcsc(t)?

Answer 17

shouldn't the characteristic equation be r^2+r=0 which gives r=0 or -1?

Answer 18

No.

Answer 19

r^2 = y''
r = y'
c = y

Answer 20

that gets me at times too

Answer 21

sorry mis-read the equation i thought the 2nd term had a prime on it - must change my glasses! :-)

Answer 22

Now, the guess. I think you mean btw, the method of undetermined coefficients.

Try yp = A sec x + B sec x.tan x + C cos x + D cos x.ln(cos x)

Answer 23

could try sin(t) for the other part .... maybe. just did a few of these before bedtime last night, but they werent triged up

Answer 24

Actually, just

yp = A sec x + B sec x.tan x + C cos x.ln(cos x)

Answer 25

where do you get those from?
just need to learn it or?

Answer 26

because it is no where to be found in my book

Answer 27

sec x comes from the in homogenous part
sec x . tan x from its derivative

Now more speculative, what happens when you try and integrate by parts tan x * something, you get something * -ln(cos x)
So sec x.ln(cos x) and cos x.ln(cos x) makes sense

Answer 28

So actually, last iteration: try

yp = A sec x + B sec x.tan x + C cos x.ln(cos x) + D sec x.ln(cos x)

Answer 29

yp = A cos(t) + B sin(t)
yp' = A' cos(t) - Asin(t) + B'sin(t) +Bcos(t)

assume: A'cos(t) + B'sin(t) = 0 cant recall y

yp' = - Asin(t) +Bcos(t)

yp'' = -A'sin(t)-Acos(t) +B'cos(t) -Bsin(t)

yp''+y = -A'sin(t)-Acos(t) +B'cos(t) -Bsin(t) +A cos(t) + B sin(t)

yp''+y = -A'sin(t)+B'cos(t) = csc(t)

Answer 30

then solve the system of equations from:

A'cos(t) + B'sin(t) = 0
-A'sin(t)+B'cos(t) = csc(t)

Answer 31

B' = csc.cos/ cos^2+sin^2 = cot

Answer 32

A' = 1/-1 = -1

Answer 33

i hope i did those crosses right

Answer 34

Ok: so what amistre's written is what I thought you meant in terms of method when you said "undefined coefficients", which goes under the usual name "Variation of parameters".

But you are using "Method of Undetermined Coefficients" which is why you need a good guess for the particular solution.

Answer 35

yeah been doing alot of this my brain is starting to go haywire lol

Answer 36

int(A')dt = int(-1)dt
A = -t


int(B')dt = int(cot(t))dt
B = ln|sin(t)|

Answer 37

So which one do you mean? Variation of Parameters or Undetermined Coefficients?

Answer 38

undetermined coefficents

Answer 39

amistre i dont see how you went from
yp = A cos(t) + B sin(t)
to that
yp' = A' cos(t) - Asin(t) + B'sin(t) +Bcos(t)
shouldnt y_p'= -aCos(t)+bSin(t)?

Answer 40

y = yh + yp

\[y=c_1cos(t)+c_2sin(t)-t\ cos(t)+ln|sin(t)|\ sin(t)\]

ill rechk it, but in the end I get this; unless im forgeting another c

Answer 41

yp = A cos(t) + B sin(t)

yp' = A' cos(t)+A cos(t)' +B' sin(t) + B sin(t)'

yp' = A' cos(t)-A sin(t) +B' sin(t) + B cos(t)

Answer 42

but A and B are constants. why would you use a' and b' ?

Answer 43

they arent constants they are unknown functions

Answer 44

we are trying to find the unknown functions A(t) and B(t)

Answer 45

amistre: you're not using the method he wants to use.

Answer 46

im aware of that ;) but its still good practice

Answer 47

well that explains why i couldn't follow along lol

Answer 48

what is the name of that method?

Answer 49

even if we use tham as variables, they are still unknowns that get 'ed

Answer 50

cant recall

Answer 51

The method amis is using is called Variation of Parameters.

I'll tell you flat out that Undetermined Coeffs is not a great method here as it's not obvious what form of the answer to use.

Answer 52

And I'll also tell you the answer so you can work towards it:

yp = cos x . ln(cos x)

Answer 53

ok. that makes so much more sense (looking at the next section)

Answer 54

midterm in 24hrs... gotta keep studying lol.

Answer 55

err, for the record, who is on the poster you're holding up in your photo?

Answer 56

osama bin ladin haha was at gun range in vegas

Answer 57

got it.

Answer 58

thanks for the help

Answer 59

i apparently cant factor r^2+1 correctly, but I can recall pretty much how to do variation param.... you know how much algebra i had to forget to learn that?