Differential Equations [Show every member is a solution]


by Jay J
Tags: differential, equations, member, solution
Jay J
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#1
Sep8-09, 02:54 PM
P: 26
problem resolved
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Mark44
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#2
Sep8-09, 03:00 PM
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Quote Quote by Jay J View Post
1. The problem statement, all variables and given/known data
PART A: For what values of r does the function y= e^ (rx) satisfy the differential equation 2y" + y' - y = 0?

PART B: If r1 and r2 are the values of r that you found in part A, show that every member of the family of functions y=ae^(r1*x) + be^(r2*x) is also a solution


2. Relevant equations
derivatives, factoring.


3. The attempt at a solution

I got part A, by taking the 1st and 2nd derivative of y and plugging into the equation and factoring to obtain r= -1 and r =1/2.

but what I'm really stuck on is how to show the answer for PART B?

Do you just plug in your r1 & r2 values from part a?

Please Help.
You should have gotten two specific values for r1 and r2, one of which happens to be a fraction. Show that the function y = aer_1 x + ber_2 x is a solution to your differential equation. That's it.
Jay J
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#3
Sep8-09, 03:02 PM
P: 26
Quote Quote by Mark44 View Post
You should have gotten two specific values for r1 and r2, one of which happens to be a fraction. Show that the function y = aer_1 x + ber_2 x is a solution to your differential equation. That's it.
I did get 2 answers, r= -1 and r=1/2. So now just plug those into the equation in part b ?

Mark44
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#4
Sep8-09, 03:04 PM
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Differential Equations [Show every member is a solution]


Quote Quote by Jay J View Post
I did get 2 answers, r= -1 and r=1/2. So now just plug those into the equation in part b ?
Yes. Then for that function, show that 2y'' + y' - y = 0.
Jay J
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#5
Sep8-09, 03:07 PM
P: 26
Quote Quote by Mark44 View Post
Yes. Then for that function, show that 2y'' + y' - y = 0.

O, okay thank you , it was really confusing me
Jay J
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#6
Sep8-09, 03:34 PM
P: 26
Quote Quote by Mark44 View Post
Yes. Then for that function, show that 2y'' + y' - y = 0.

When I Plugged back into 2y'' + y' -y = 0 I get 2ae-x+1/2be1/2x -ae-x +1/2be1/2x -ae-x -be1/2x=0

and theres nothing to cancel out ?

Help?
Mark44
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#7
Sep8-09, 03:51 PM
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Quote Quote by Jay J View Post
When I Plugged back into 2y'' + y' -y = 0 I get 2ae-x+1/2be1/2x -ae-x +1/2be1/2x -ae-x -be1/2x=0

and theres nothing to cancel out ?

Help?
2ae-x+1/2be1/2x -ae-x +1/2be1/2x -ae-x -be1/2x = e-x(2a - a - a) + e.5x(.5b + .5b - b) = ?
Jay J
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#8
Sep8-09, 03:53 PM
P: 26
Quote Quote by Mark44 View Post
2ae-x+1/2be1/2x -ae-x +1/2be1/2x -ae-x -be1/2x = e-x(2a - a - a) + e.5x(.5b + .5b - b) = ?
ahh totally 4got about that ex . . lets see what I can do now lol
Jay J
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#9
Sep8-09, 08:25 PM
P: 26
um I am still having some trouble with this . . cause even after factoring it your left with 2 unknown variables so how are you to prove it?
Jay J
Jay J is offline
#10
Sep8-09, 09:00 PM
P: 26
Never Mind, I'm an idiot I already answered the Question lmao

Thanks Guys.


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