Question

can someone help with 2nd order ODE homogeneous equation..

Answer 1

2x"-7x'+6x=0

x'=dx/dt
x"=d^2/dt^2

Answer 2

char equation
is
2r^2-7r+6=0
so r=1.5 ,2

now
x = a exp(1.5 t ) +b exp(2 t)

Answer 3

i got to that part...and I get a and b equal zero...where am I going wrong?

Answer 4

initial conditions...

Answer 5

x(0)=0
x'(0)=0

Answer 6

apply the initial condition

Answer 7

x = a exp(1.5 t ) +b exp(2 t)

x'=1.5 a exp(1.5 t) +2b exp(2 t)

a,b goes to zero
are you sure about initial conditions?

Answer 8

yes and looking back, I just remembered the professor asking us what is so particular about them both equaling zero...what does it mean?

Answer 9

nevermind, classmate has these IC:

x(0)=1
x'(0)=0

but now I get -B+B=1...doesn't make sense

Answer 10

what grade r u in ?
is it for ms degree ode ?

Answer 11

senior engineer undergrad

Answer 12

what's the question?

Answer 13

x(0)=1
x'(0)=0
are those initial condition ?

Answer 14

^ whatever he said

Answer 15

@amoodarya yes

Answer 16

because after you find your x = a exp(1.5 t ) +b exp(2 t)
you have to apply x(0)= 1 first
then take the derivative of x = a exp(1.5 t ) +b exp(2 t)
and apply x'(0) =0

Answer 17

x = a exp(1.5 t ) +b exp(2 t) x(0)=1 --> a+b=1

x'=1.5 a exp(1.5 t) +2b exp(2 t) x'(0)=0 --> 1.5a +2b=0
now find a,b

Answer 18

I normally use c1 and c2 for these type of problems... easier to see
x = c1e^1.5t + c2e^2t

Answer 19

thank you, i'm getting all my writings mixed up

Answer 20

so I assume that I'm dealing with a x(t) = c1e^1.5t + c2e^2t
x(0) = 1
1 = c1+ c2

Answer 21

e^ 0 is 1 ^^ but multiplied by c1 and it's c1

Answer 22

so we need to product rule for x(t)

Answer 23

a+b=1
1.5a+2b=0

if u solve
a=4 , b=-3

Answer 24

let me double check that
x(t) = c1e^1.5t + c2e^2t
x'(t) = c1 1.5e^1.5t + c2 2e^2t

Answer 25

x'(0) = 0
0 = 1.5 c1 +2c2

Answer 26

1 = c1+ c2
0 = 1.5 c1 +2c2

Answer 27

subsitution method... or matrix method.. would be easier.

Answer 28

@amoodarya can you help with the non homogeneous part now?

Answer 29

yes

Answer 30

ok I will post in a bit and tag you..

Answer 31

uh oh... use subsitution method to solve for c1 and c2.

Answer 32

4 and - 3 is right

Answer 33

ok so the whole problem was..
2x"-7x'+6x=sin2t
i solved for the homogeneous part, now i need to solve for the particular solution..the professor showed us how to do it if the right side was a constant but you can see that it is not so I'm not sure what I'm suppose to do now @amoodarya

Answer 34

oh these.. undetermined coefficents.

Answer 35

yes you do have to find the Yh first which is what we did... we just need Yp

Answer 36

thank you for your input also. it sounds familiar..

Answer 37

there's a format for the right hand side... there's no e^xsinx or e^x cosx so it may make life easier...
sinwx coswx has the format of Asinwx + Bcoswx where w = the number

Answer 38

sin2t = Asin2t + Bcos2t

Answer 39

but they are new constants?

Answer 40

Yp = Asin2t + Bcos2t
find the first and second derivatives of this equation... then plug it in to 2x"-7x'+6x=sin2t
now if you have a 0 = sin 2t that means that our initial Yp fails and we need to add a mulitplier

Answer 41

so find the derivatives of Yp and plug it in to the equation

Answer 42

then collect all of your A's and B's.

Answer 43

you mean plug it in x" and x' after I find the derivatives?

Answer 44

:) yes

Answer 45

ok

Answer 46

so far I have -74sin2t-32cos2t+24e^(3/2t)-18e^(2t)=sin2t

Answer 47

gotta go for a bit brb

Answer 48

ummm whoa no no no no no no no no no that's not what I meant X*(

Answer 49

you have to plug in the derivatives and the original Yp in 2x"-7x'+6x=sin2t
and I recommend that you use y instead
2y''-7y'+6y = sin2t

Answer 50

o oops

Answer 51

I'm following...but i forgot to ask why sin2t=Asin2t+Bcos2t? just a rule i need to remember?

Answer 52

yeah it's a format.... you need to remember
for sinwt and coswt

it's Asinwt+Bcoswt where w = the number your assigned if applicable.

Answer 53

there's four of them...
e^cx c = exponent number
x^n n= degree number
the one I typed earlier and another one with the one I typed earlier with e^cx