Question

Solve the initial value problem for x as a function of t.
(t^2 -3t+2)dx/dt =1 (t>2), x(3)=0

Answer 1

integrate it?

Answer 2

i have no idea. i am lost with this one

Answer 3

solve for dx/dt

Answer 4

and integrate

Answer 5

simply dvide both ides by the (t^2-3t+2), then integrate both sides

Answer 6

you will use partial fractions here

Answer 7

that will get you up to x family then just use the x(3) to find the exact function

Answer 8

if asnaseer dosent solve it for you i will

Answer 9

ok :)

Answer 10

\[(t^2-3t+2)*\frac{dx}{dt}=1\]so:\[\int dx=\int \frac{dt}{t^2-3t+2}\]

Answer 11

yes very nice^

Answer 12

this will give you:\[x=\ln(\frac{2-t}{1-t})+c\]then use the initial values you were given to find c.

Answer 13

ok. thank you very much asnaseer

Answer 14

i think there should be another part though, check your answer book blue12

Answer 15

yw - @Outkast3r09 basically gave the same steps but just in words.

Answer 16

no i dont see another part

Answer 17

well, maybe im wrong

Answer 18

it is page 462. problem 51.

Answer 19

you'l need to use partials to solve

Answer 20

i believe he's right

Answer 21

dont worry about it the intal values give you c

Answer 22

yeah he is right. @Outkast3r09

Answer 23

hm alright@ LagrangeSon678

Answer 24

i am stuck again!. i got x=ln(2-t)/(1-t)+C
how do i find C now?

Answer 25

i know i am dumb..

Answer 26

hy Outkast3r09 can u plz explain

Answer 27

you have to put in the values for x(t)=0 you're going to get imaginary numbers i believe since inside the natural log will be negatives

Answer 28

now, replace t with 0

Answer 29

remeber ln(1)=0

Answer 30

t with 0??

Answer 31

in your anti derivaitve

Answer 32

yes but i think the integration is off?

Answer 33

o no its good

Answer 34

why are you replacing t with 0... shouldn't it be replaced with 3?

Answer 35

x=ln(t-2)-ln(t-1)+ln2 is the answer in the book

Answer 36

o wait nvm i thought it was lnor ln

Answer 37

ln(1/2)+c=0
c=-ln(1/2)

c=-ln(1)-(-ln(2)= 0+ln(2)

Answer 38

ln(t-2)-ln(t-1)=ln(t-2)/ln(t-1)

Answer 39

so all together you get
\[\frac{\ln(t-2)}{\ln(t-1)}+\ln(2)\]

Answer 40

hm i get it thanku OutKast3r09 and LagrangeSon678

Answer 41

anybody know about parametric equation

Answer 42

that would hav ebeen insane if it was
\[\frac{\ln(t-2)}{ln(t-1)}\]

I was sitting here going .... hmm how would this be possible without imaginary... and i know smoewhat lagrange

Answer 43

i know a lil about parametric equation

Answer 44

can i get some help

Answer 45

with?

Answer 46

Find the parametric equations and a parameter interval for the motion of a particle that starts out at (a,0) and traces the x^2+y^2=a^2

Answer 47

well the interval would be 0,infty due to squared i believe

Answer 48

a^2 is a constant i belive lagrange?

Answer 49

This is an equation of a circle with radius a.