1. Homog Lin... #2

2.) Find the general sol'n of hte following Dif EQ: $\frac{d^5y}{dx^5}-2\frac{d^4y}{dx^4} + 17\frac{d^3y}{dx^3} = 0.$

Work:

So, $m^5 - 2m^4 + 17m^3 = 0$

Factor: m^3*(m^2 - 2m + 17)

Thus, m = 0.

So, the general solution is just $y = c_1 + c_2$ since e^0 = 1 just cancels out.

2. Hello, Ideasman!

Your algebra is off . . .

2) Find the general sol'n of: . $\frac{d^5y}{dx^5}-2\frac{d^4y}{dx^4} + 17\frac{d^3y}{dx^3} \:= \:0$

We have: . $m^5 - 2m^4 + 17m^3\:=\:0\quad\Rightarrow\quad m^3(m^2-2m + 17) \:=\:0$

We have two equations to solve:

. . $m^3\,=\,0\quad\Rightarrow\quad m\,=\,0\quad\Rightarrow\quad y \:=\:C_1$

. . $m^2-2m + 17\:=\:0\quad\Rightarrow\quad m \:=\:\frac{2 \pm\sqrt{-64}}{2} \:=\:1 \pm 4i \quad\Rightarrow\quad y \:=$ . $e^x\left(C_2\cos4x + C_3\sin4x\right)$

General solution: . $y \;=\;C_1 + e^x\left(C_2\cos4x + C_3\sin4x\right)$

3. Originally Posted by Soroban
Hello, Ideasman!

Your algebra is off . . .

We have: . $m^5 - 2m^4 + 17m^3\:=\:0\quad\Rightarrow\quad m^3(m^2-2m + 17) \:=\:0$

We have two equations to solve:

. . $m^3\,=\,0\quad\Rightarrow\quad m\,=\,0\quad\Rightarrow\quad y \:=\:C_1$

. . $m^2-2m + 17\:=\:0\quad\Rightarrow\quad m \:=\:\frac{2 \pm\sqrt{-64}}{2} \:=\:1 \pm 4i \quad\Rightarrow\quad y \:=$ . $e^x\left(C_2\cos4x + C_3\sin4x\right)$

General solution: . $y \;=\;C_1 + e^x\left(C_2\cos4x + C_3\sin4x\right)$

Thank you Soroban. I see how I was careless in computing the root for m^2 - 2m + 17!