Differential Equations [Show every member is a solution]

In summary, the problem was solved by finding the values of r1 and r2 that satisfy the differential equation 2y'' + y' - y = 0.
  • #1
Jay J
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problem resolved
 
Last edited:
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  • #2
Jay J said:

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 = aer_1 x + ber_2 x is a solution to your differential equation. That's it.
 
  • #3
Mark44 said:
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 ?
 
  • #4
Jay J said:
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.
 
  • #5
Mark44 said:
Yes. Then for that function, show that 2y'' + y' - y = 0.


O, okay thank you , it was really confusing me :confused:
 
  • #6
Mark44 said:
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 there's nothing to cancel out :uhh:?

Help?
 
  • #7
Jay J said:
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 there's nothing to cancel out :uhh:?

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) = ?
 
  • #8
Mark44 said:
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 . . let's see what I can do now lol
 
  • #9
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?
 
  • #10
Never Mind, I'm an idiot I already answered the Question lmao

Thanks Guys.
 

1. What are differential equations?

Differential equations are mathematical equations that describe how a variable changes over time, taking into account the rate of change of that variable. They are used to model a wide range of phenomena in science and engineering, from population growth to the motion of planets.

2. How do you solve differential equations?

The process of solving a differential equation involves finding a function or set of functions that satisfy the equation. This can be done analytically, using mathematical techniques such as separation of variables or integrating factors, or numerically using computer algorithms.

3. What is the difference between ordinary and partial differential equations?

Ordinary differential equations involve a single independent variable, while partial differential equations involve multiple independent variables. Ordinary differential equations are typically used to model phenomena that change over time, while partial differential equations are used to model phenomena that vary in multiple dimensions.

4. How do you know if a function is a solution to a differential equation?

In order for a function to be a solution to a differential equation, it must satisfy the equation for all values of the independent variable. This means that when the function is plugged into the equation, the equation must be true. Additionally, the solution must also satisfy any initial conditions or boundary conditions specified in the problem.

5. Can every member of a set be a solution to a differential equation?

No, not every member of a set will necessarily be a solution to a differential equation. The equation must be satisfied for all values of the independent variable, and not all functions will meet this requirement. However, there may be a set of functions that are all solutions to a particular differential equation.

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