# Thread: Solving a fifth order differential equation?

1. ## Solving a fifth order differential equation?

Solve: y'''''-3y''''-y'''+23y''-64y'+60y=x^4 + e^2x + e^3x + sinx; when x=0, y=a, y'=a2, y''=a3, y'''=a4, y''''=a5

I wanted to approach this the way I would a second order, ie. y= yc+yp. would that be the way to go? if so, how would i factor the left side? thanks in advance

2. Since they're constant coefficient, I assume that process should be fine.

Assume a solution of the form $\displaystyle y = e^{rx}$, then for the solution to the homogeneous equation, you would have

$\displaystyle r^5e^{rx} - 3r^4e^{rx} - r^3e^{rx} + 23r^2e^{rx} - 64r\,e^{rx} + 60e^{rx} = 0$

$\displaystyle r^5 - 3r^4 - r^3 + 23r^2 - 64r + 60 = 0$.

Now use the factor theorem and long division to factorise this, giving

$\displaystyle (r - 2)^2(r + 3)(r^2 - 2r + 5) = 0$.

Therefore $\displaystyle r = 2$ (multiplicity 2), $\displaystyle r = -3$, $\displaystyle r = 1 - 2i$, $\displaystyle 1 + 2i$.

So the homogeneous solution is

$\displaystyle y_c = Ae^{2x} + Bx\,e^{2x} + Ce^{-3x} + De^{x}(E\cos{2x} + F\sin{2x})$.

3. Try the factors of 60. Use synthetic division to check.

[EDIT]: ProveIt beat me to it.

4. Thanks a lot you guys! This was a huge help!

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# how to solve fifth order differential equations

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