Question

How do you find the appropriate table for the differential equation dy/dx= x/x-2?

Answer 1

I can solve your equation

Answer 2

okay. Is that what I need to do first?

Answer 3

yes! I think so! I f we know the formula for y(x), then we can compute all its values, each for each value of x

Answer 4

I'm not sure how to do this type of problem. I'm use to them having at least one y in them.

Answer 5

do you know how to solve your differential equation?

Answer 6

no, not really how to begin

Answer 7

it is very simple, we have to solve this integral:
\[y\left( x \right) = \int {\frac{x}{{x - 2}}} \;dx\]

Answer 8

2ln(x-2)+x+C ?

Answer 9

yes!

Answer 10

so do i just fill in values for x?

Answer 11

a better expression is:
\[y\left( x \right) = x + \ln \left[ {{{\left( {x - 2} \right)}^2}} \right] + C\]
Now we have to keep in mind that the solution y(x) exists for all x such that:
\[x \ne 2\]

Answer 12

its not really working for what i need. I need the values of 0 and .5

Answer 13

then we can try to insert into that formula several values of x and consider the corresponding values of y

Answer 14

ok! Then we can replace x with 0 and x with 1/2, what values of y do you get?

Answer 15

we can neglect the arbitrary constant C

Answer 16

what is y(0)=...?
what is y(0.5)=...?

Answer 17

0 was undefined and .5 was 1.3...

Answer 18

those answers weren't right for what they gave

Answer 19

sorry for x=0 we have:
y(0) = 0+ ln 4= 2*ln2

Answer 20

maybe we have to compute x, when y=0 and y= 0.5?

Answer 21

zero has to either be undefined or 0 and . 5 has to be -1 or -1/3

Answer 22

@Michele_Laino

Answer 23

I'm trying to choose an appropriate value of the arbitrary constant

Answer 24

if we set C=-2ln 2, then we get:
y(0) = 0

Answer 25

would the one equal -1/3 at that point?

Answer 26

please wait a moment!

Answer 27

okay thanks for all your help btw!

Answer 28

after we set C=-2 ln2= -ln4, we get:
\[y\left( x \right) = x + \ln \left[ {{{\left( {x - 2} \right)}^2}} \right] - \ln 4\]

Answer 29

and y(0.5) is:
\[y(1/2) = \frac{1}{2} + \ln \left[ {{{\left( {\frac{1}{2} - 2} \right)}^2}} \right] - \ln 4 = - 0.076\]

Answer 30

okay thank you!

Answer 31

thank you!