Question

How to solve: x'=y, y'=x? Please, help

Answer 1

Won't x' be 1 ?

Answer 2

Nope

Answer 3

o.O Differentiating with respect to sme other variable?

Answer 4

The result from the book says; \(x(t) = C_1e^t-C_2e^{-t}= y(t)\)

Answer 5

can't you do x'=y so x''=y'

and then substitute
x''=x

Answer 6

\[x''-x=0 \\ r^2-1 =0 \text{ is characteristic equation }\]

Answer 7

sorry, \(y(t) = C_1e^t+C_2e^{-t}\)

Answer 8

Show me, please. @freckles

Answer 9

show you what?

Answer 10

\[r=\pm 1 \]

Answer 11

How to doctor up the problem

Answer 12

\[x(t)=a e^t+b e^{-t}\]

Answer 13

and then you can use that x'=y

Answer 14

differentiate that to get y

Answer 15

To me, I do :
\(\dfrac{dx}{dt}= y\) that gives me \(x(t) = yt +C\)
Same as \(\dfrac{dy}{dt}= x\) hence \(y(t) = xt + A\)
then I am stuck, hehehe

Answer 16

did you understand my way ?

Answer 17

like we have
x'=y, y'=x



so if x'=y then x''=y'

replace the y' in the second equation with x''

Answer 18

you get x''=x

Answer 19

I think I got it, Let me show
y' =x hence x' = y"

Answer 20

hence y"=y or y"-y= 0 and solve for y, right?

Answer 21

you can differentiate the other equation which is what you have done great

Answer 22

Ok, I think I got it. Thanks a ton. BTW, do you know partial differential equation? I need help on it also. :)

Answer 23

i know some stuff
other stuff no

like asking me is like a 30/70 chance of me knowing
like the 30 me knowing
the 70 me not knowing
:p

Answer 24

I think there is another way to solve that system you mentioned above by the way

Answer 25

I could be wrong does it involve matrices and eigenvalues ?

Answer 26

yup, may be

Answer 27

but that isn't the way you were trying to solve?

Answer 28

I don't care, as long as it makes sense to me, I am ok with any method

Answer 29

k k