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

[Saladsamurai]

Homework Statement

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!

 

[Saladsamurai]

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.

 

[cristo]

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

 

[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.

 

[Saladsamurai]

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.

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

 

[rock.freak667]

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

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

 

[D H]

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

$$\aligned<br /> C_1 &= c_3 \\<br /> C_2 &= c_4 \\<br /> C_3 &= c_5 \\<br /> C_4 &= c_1 - 3c_2 \\<br /> C_5 &= c_2<br /> \endaligned$$

 

[Saladsamurai]

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

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.

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

$$\aligned<br /> C_1 &= c_3 \\<br /> C_2 &= c_4 \\<br /> C_3 &= c_5 \\<br /> C_4 &= c_1 - 3c_2 \\<br /> C_5 &= c_2<br /> \endaligned$$

Ok! Thanks guys

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

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

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?

 

[rock.freak667]

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?

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.

 

[Saladsamurai]

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.

Ah ha. I see. I was trying to plug in my general solution to

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

which will of course have more terms than the general solution to just

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

Thanks again!