# Diff Eq: Is it me? Or Wolfram? I know, it's me.

1. Sep 27, 2010

1. The problem statement, all variables and given/known data

y(5) - 2y(4) + y(3) = 0

This is 5th-order, linear, HG, with constant coefficients, and hence I used the characteristic equation:

m5 - 2m4 + m3 =

m3(m - 1)(m - 1) = 0

Hence, m = {1,1,0,0,0} and so the solution is given by:

y(x) = C1 + C2*x +C3*x2 + C4*exp(x) + c5*x*exp(x).

This is the Wolfram solution.

They seem to have an extra exponential term inside of the parentheses. Am I missing something?

Thanks!

2. Sep 27, 2010

Hmmm.... looking at the characteristic EQ, it would seem that:

m3(m2 - 2m + 1) = 0 and hence:

(m2 - 2m + 1) = 0 and so

y'' - 2y' + y = 0

But when you look at my solution, differentiation will cause my C1 term to drop out of the derivatives altogether, and hence it will not be able to cancel.

I have messed up somehow, but cannot seem to rectify it. I have tunnel vision right now and cannot seem to convince myself that my method is not correct.

3. Sep 27, 2010

### cristo

Staff Emeritus
Doesn't the wolfram solution reduce to yours on setting their c_1-3c_2 equal to your C_4?

4. Sep 27, 2010

### rock.freak667

In Wolfram's answer, you can just write c1-3c2 as some other symbol and get your same answer. But I am not sure exactly how Wolfram solved it.

5. Sep 27, 2010

I am still not seeing how that rectifies this. Maybe I need a coffee ...

6. Sep 27, 2010

### rock.freak667

Your solution would not solve y'' - 2y' + y = 0

it will solve d3/dx3 (y'' - 2y' + y) to give 0.

7. Sep 27, 2010

### D H

Staff Emeritus
Your solution (I changed c5 to C5):

$$y(x) = C_1 + C_2x +C_3x^2 + C_4\exp(x) + C_5x\exp(x)$$

Wolfram's solution:

$$y(x)=c_5x^2 + c_4x+e^x(c_2x+c_1-3c2)+c3$$

Those solutions are the same. To see this, set

\aligned C_1 &= c_3 \\ C_2 &= c_4 \\ C_3 &= c_5 \\ C_4 &= c_1 - 3c_2 \\ C_5 &= c_2 \endaligned

8. Sep 27, 2010

Ok! Thanks guys

Hmmm.

$$\frac{d^5y}{dx^5} - 2\frac{d^4y}{dx^4} + \frac{d^3y}{dx^3} = 0$$

$$\Rightarrow \frac{d^3y}{dx^3}\left(\frac{d^2y}{dx^2} - 2\frac{dy}{dx} +1\right) = 0$$

Ok. I see that it wil not solve y'' - 2y' + y = 0. But from the above, should it not solve

$$\left(\frac{d^2y}{dx^2} - 2\frac{dy}{dx} +1\right) = 0$$

which does not look promising. I know that I am seriously abusing the differential operator, but it should work eh?

9. Sep 27, 2010

### rock.freak667

y'' - 2y' + y = 0

has a solution of y=(Ax+B)e-x, which will solve y'' - 2y' + y = 0

Your general solution which contains more terms, I do not think will solve y'' - 2y' + y = 0.

10. Sep 27, 2010

$$\frac{d^5y}{dx^5} - 2\frac{d^4y}{dx^4} + \frac{d^3y}{dx^3} = 0$$
$$\left(\frac{d^2y}{dx^2} - 2\frac{dy}{dx} +1\right) = 0$$