Differential Equations [Show every member is a solution]

by Jay J
Tags: differential, equations, member, solution
 P: 26 problem resolved
Mentor
P: 19,753
 Quote by Jay J 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.
P: 26
 Quote by Mark44 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 ?

Mentor
P: 19,753

Differential Equations [Show every member is a solution]

 Quote by Jay J 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.
P: 26
 Quote by Mark44 Yes. Then for that function, show that 2y'' + y' - y = 0.

O, okay thank you , it was really confusing me
P: 26
 Quote by Mark44 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?
Mentor
P: 19,753
 Quote by Jay J 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) = ?
P: 26
 Quote by Mark44 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
 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?
 P: 26 Never Mind, I'm an idiot I already answered the Question lmao Thanks Guys.

 Related Discussions Calculus & Beyond Homework 4 Calculus & Beyond Homework 2 Calculus & Beyond Homework 2 Calculus & Beyond Homework 11 Physics Learning Materials 0