Question

Can someone please help!! Find the general solution of the system of differential equations

x'=10x−14y
y'=25x−35y
(where primes indicate derivatives with respect to t) by using the initial conditions
x(0)=A y(0)=B

x(t)=
y(t)=

Answer 1

This is what I got so far

Answer 2

on the 2nd column of your work, you didn't distribute a negative
the x's should cancel leaving
x'' +25x' = 0

Answer 3

so r would then equal +/- 5i and then what would we do?

Answer 4

and thanks sooo much for replying!

Answer 5

wait no, r = 0,-25

r^2 +25r = 0
r(r+25) = 0

Answer 6

ok but what do we do with those values to get x(t) and y(t)?

Answer 7

you can form the general solution for x(t)

--> x(t) = c1e^-25t + c2

then you have to repeat this process to find y(t)...substitute to get an equation with y'',y' and y

this is a great resource
http://tutorial.math.lamar.edu/Classes/DE/RealRoots.aspx

Answer 8

if x(0)=A then A=c1+c2 right

Answer 9

correct, looks like you can't find the constants until after you have y(t)

Answer 10

but how do we solve for y(t)? we knowvthat y'=25x-35y and y=-x'-10x/14 based on the first equation

Answer 11

same way we solved for x(t)
take 2nd equation, solve for x
differentiate and set equal to 1st equation

Answer 12

so you end up getting y'=25x-35y
x=y'+35y/25
x'=y''+35y'/25
x'=10x-14y
y''+35y'/25=10(y'+25y/25)-14y

Answer 13

right?

Answer 14

do e plug in the y we found earlier?

Answer 15

we*

Answer 16

without adding in the value of y we found earlier i ended up with y''+25y'=0

Answer 17

looks good...if you continue you will find you get same answer as before
y'' +25y' = 0

--> r =0,-25

Answer 18

so now
y(t) = k1e^-25t +k2
where
k1+k2 = B

Answer 19

but you then get B=c1+c2

Answer 20

i used k just to distinguish the constants for x and y
im working on it
it seems the hard part is solving for the constants

Answer 21

ok

Answer 22

so they want the answer in terms of A and B right?

Answer 23

I believe so

Answer 24

do we use determinants?

Answer 25

i don't think so.

ok by differentiating we can find x'(0) and y'(0)
x'(0) = -25c1
y'(0) = -25k1
Now from original equations and that x(0) = A and y(0) = B
-25c1 = 10A -14B
-25k1 = 25A-35B

thus we can now solve for c1 and k1
then find c2,k2 by using fact that
c2 = A-c1
k2 = B-k1

Answer 26

so x(t)=-c2-Ae^-25t+A-c1

Answer 27

can't have any c's in the answer, only A and B

Answer 28

but then what ould it be

Answer 29

would*

Answer 30

x'(0) = 10x(0) -14y(0)
-25c1 = 10A-14B
now solve for c1 by dividing by -25
do you see it now

Answer 31

sort of

Answer 32

x(t)=((10A-14B)/-25)e^(-25t)+A-(10A-14B/-25)

Answer 33

correct...you can simplify the fractions a bit

Answer 34

do same thing with y
y'(0) = 25x(0) -35y(0)

Answer 35

but for x(t) the answer ended up being wrong

Answer 36

oh do you have the answer

Answer 37

no its from an online homework set and it only tells u if yr wrong or right

Answer 38

you may have needed to simplify it...did you enter it same as you wrote above?

Answer 39

yea i did but it states that u dont need to simplify it

Answer 40

http://www.wolframalpha.com/input/?i=x%27%28t%29+%3D+10x%28t%29-14y%28t%29%2C+y%27%28t%29%3D25x%28t%29-35y%28t%29%2C+x%280%29%3DA%2C+y%280%29%3DB

i checked with wolfram, i am correct
look at an example so you know what exact form your answer should be in

Answer 41

simplified i get
\[x(t) = \frac{14B-10A}{25}e^{-25t} + \frac{35A-14B}{25}\]

Answer 42

Note: wolfram puts answer in a complicated form

Answer 43

ok thank you

Answer 44

thanks a bunch!

Answer 45

your welcome