# Thread: Need help with elimination method

1. ## Need help with elimination method

Okay, I need a plain english explaination of how to go about solving problems like this. This one was given as an example in my textbook but the author skipped too many steps, so you won't spoil anything by answering it.
Using the elimination method, find a general solution where differentiation is with respect to t.

x' = 3y
y' = 2x - y

2. Originally Posted by diddledabble
Okay, I need a plain english explaination of how to go about solving problems like this. This one was given as an example in my textbook but the author skipped too many steps, so you won't spoil anything by answering it.
Using the elimination method, find a general solution where differentiation is with respect to t.

x' = 3y
y' = 2x - y
Differentiate the second equation again:

y''=2x'-y'

Now substitute from the first into this to get:

y''=6y-y'

which is a second order constant coefficient linear homogeneous ODE and hence can be solved with no trouble.

Solve it, now substitute the solution back into the second equation y'=2x-y to find x.

CB

3. ## Check my work

Okay, so from what Captain said and using my book I got this...
Auxiliary equation
r^2+r-6=0
(r+3)(r-2)=0 so r= -3, 2

y(t)=c1e^(-3t)+c2e^(2t)
y'(t)=-3c1e^(-3t)+2c2e^(2t)

Subbing back into y'=2x-y
-2c1e^(-3t)+3c2e^(2t)=2x
x(t)=-c1e^(-3t)+ (3/2)c2e^(2t)

Also if someone can type these in LaTex I would appreciate it. I tried and it didn't work out very well. Thanks!

4. Originally Posted by diddledabble
Okay, so from what Captain said and using my book I got this...
Auxiliary equation
r^2+r-6=0
(r+3)(r-2)=0 so r= -3, 2

y(t)=c1e^(-3t)+c2e^(2t)
y'(t)=-3c1e^(-3t)+2c2e^(2t)

Subbing back into y'=2x-y
-2c1e^(-3t)+3c2e^(2t)=2x
x(t)=-c1e^(-3t)+ (3/2)c2e^(2t)

Also if someone can type these in LaTex I would appreciate it. I tried and it didn't work out very well. Thanks!
Translated into LaTeX:

Okay, so from what Captain said and using my book I got this...
Auxiliary equation

$r^2+r-6=0$
$(r+3)(r-2)=0$ so $r= -3, 2$

$y(t)=c_1\;e^{-3t}+c_2\;e^{2t}$
$y'(t)=-3c_1\;e^{-3t}+2c_2\;e^{2t}$

Subbing back into $y'=2x-y$

$-2c_1\;e^{-3t}+3c_2\;e^{2t}=2x$
$x(t)=-c_1\;e^{-3t}+ (3/2)c_2\;e^{2t}$

Also if someone can type these in LaTex I would appreciate it. I tried and it didn't work out very well. Thanks!

CB