# Given the solution find the differential equation

1. Nov 21, 2015

### JoshW

1. The problem statement, all variables and given/known data
For each Set write the differential equation they are a solution to. It is also asked to find the wronskian but I remember how to do that.
(I am doing this on my own to brush up before I have to take mathphys next semester)
1. e-x,e-4x
2, 1,x,x2
2. Relevant equations

3. The attempt at a solution
1.(1+1/16)D"=e-x+e-4x
y"+y'+y=x2+x+1 but that doesn't seem right

If someone could really explain how to do this or walk me through it I would be very happy!

2. Nov 21, 2015

### Staff: Mentor

Correct, it's not right. How did you get 1 + 1/16?

I'm hopeful that you still have your textbook from when you studied diff. equations. Some things are 1) the characteristic equation for a diff. equation, and 2) the Wronskian.

For the first problem, since you have two distinct exponentials, the diff. equation will be 2nd order, homogeneous, with constant coefficients.
For the second, since there are three solutions, the diff. equation will be 3rd order, homogeneous, with constant coefficients as well. The fact that the solutions are 1, x, and x2 is a clue to the number of distinct roots of the characteristic equation.

3. Nov 21, 2015

### JoshW

The book we used was Boas. So we may as well have not had a textbook at all.

So I need to find the roots to the equation? I am so lost.

so for the first my equation will be y"+y?

and the second I was correct? or for the second is it just a0x2+a1x+a2

I apologize I have trouble understanding what I read. I suffered head trauma as a kid and well I have trouble understanding anything unless it is spoken to me.

4. Nov 22, 2015

### Staff: Mentor

The roots of the characteristic equation. For example, the differential equation y' - 3y = 0 has a characteristic equation of r - 3 = 0.

The characteristic equation is obtained by assuming there is a solution of the form $y = e^{rx}$
Differentiating, we get $y' = re^{rx}$
Substituting into the diff. equation, we have
$y' - 3y = 0$, and $y = e^{rx}$
$\Rightarrow re^{rx} - 3e^{rx} = 0$
$\Rightarrow (r - 3)e^{rx} = 0$
So r - 3 = 0 is the characteristic equation for the diff. equation y' - 3y = 0
The only root of this char. equation is r = 3, which gives us a solution of $y = e^{3x}$
The general solution of y' - 3y = 0 is $y = c_1e^{3x}$.

Another form of notation for the diff. equation uses the differentiation operator D. In this notation, the diff. equation would be (D - 3)y = 0. This notation is useful as you can pick off the characteristic equation directly.
No, for two reasons.
1. y'' + y is NOT an equation.
2. If the diff. eqn were y'' + y = 0, the characteristic equation would be $r^2 + 1 = 0$.
No. At first I didn't think you had even done any work for #2 -- you didn't give any indication that your work was for problem 2.

No. The problem is to find a differential equation. The above is not an equation, nor is it anywhere close to what the differential equation would be whose solutions are 1, x, and x2.

You have a lot to catch up on. Let's focus on the first question before tackling the second one.

Last edited: Nov 24, 2015
5. Nov 23, 2015

### JoshW

right ok. So for the first:
y=e-4x
y'=-4e-4x
y''=16e-4x

/∴ (y"-16y)=0 if y=e-4x
same thing for y=e-x

(y"-y)=0

I am not sure If I have to combine the solutions to form something else

for #2
y'=2x
y"=2

/∴ (½y"+½y'+y)=0

6. Nov 24, 2015

### Staff: Mentor

No.
For the equation you came up with, the fundamental solutions are $y = e^{-4x}$ and $y = e^{4x}$
The characteristic equation of your diff. equation is $r^2 - 16 = 0$ or $(r - 4)(r + 4) = 0$
You want a diff. equation whose characteristic equation has roots of -1 and -4, not 4 and -4.

Again, no.
Here the characteristic equation is $r^2 - 1 = 0$ or (r - 1)(r + 1) = 0, so the solutions are r = 1, or r = -1.
No. Again, let's hold off on this one until you get squared away on the easier one, above.

7. Nov 24, 2015

### JoshW

How do I know that i want -1 and -4 as the roots?

8. Nov 24, 2015

### Staff: Mentor

Because the two functions below are solutions to the diff. equation.

9. Nov 25, 2015

### JoshW

I appreciate the help! but this really isn't going anywhere. So thank you I will just wait untill DE's are introduced in a class.