Differential Equations [Show every member is a solution]

[Jay J]

problem resolved

 

[Mark44]

Homework Statement

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

Homework Equations

derivatives, factoring.

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 = ae^{r_1 x} + be^{r_2 x} is a solution to your differential equation. That's it.

 

[Jay J]

You should have gotten two specific values for r₁ and r₂, one of which happens to be a fraction. Show that the function y = ae^{r_1 x} + be^{r_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]

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]

Yes. Then for that function, show that 2y'' + y' - y = 0.

O, okay thank you , it was really confusing me

 

[Jay J]

Yes. Then for that function, show that 2y'' + y' - y = 0.

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

and there's nothing to cancel out ?

Help?

 

[Mark44]

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

and there's nothing to cancel out ?

Help?

2ae^-x+1/2be^1/2x -ae^-x +1/2be^1/2x -ae^-x -be^1/2x = e^-x(2a - a - a) + e^.5x(.5b + .5b - b) = ?

 

[Jay J]

2ae^-x+1/2be^1/2x -ae^-x +1/2be^1/2x -ae^-x -be^1/2x = e^-x(2a - a - a) + e^.5x(.5b + .5b - b) = ?

ahh totally 4got about that e^x . . let's see what I can do now lol

 

[Jay J]

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]

Never Mind, I'm an idiot I already answered the Question lmao

Thanks Guys.