Solving the Equation: k^3-3K2+31k-37=0 with Synthetic Division

[skysurani]

solve the following promblem
Y^(,,,)-3y^(,,)+31y^(,)-37y=0
i let y = e^kx
y'= ke^kx
y''=k^2e^kx
y'''=k^3e^kx

so i got this
k^3e^kx-3(k^2e^kx)+31(ke^kx)-37(e^kx)=0
e^kx(k^3-3K2+31k-37)=0

so,

(k^3-3K2+31k-37)=0

now i have to find (K)


how should i solve for (k) from this equation k^3-3K2+31k-37=0
can i use synthetic division if yes how should i use it or which other method can i use

is this right

i solve the k by synthetic division
5 1 1 -17 -65
5 30 65
1 6 13 0

so the factor is (k-5) (K^2+6k+13)
then i use this equation
(-b+-squrt(b^2-(4ac)))/2a

and got k = -3 +- 2i

and my fianl answer is
y=c1e^5x+c2e^(-3x)cos(2x)+c3e^(-3x)sin(2x)

 

[dextercioby]

Is this the equation

$$\frac{d^{3}y}{dx^{3}}-3\frac{d^2y}{dx^{2}}+31\frac{dy}{dx}-37y=0$$

Daniel.

 

[graphic7]

Since the equation is homogeneous and has constant coefficients, you can just solve its characteristic equation.

You'll have an equation of the form:

$$y(x) = c_1 e^{\lambda_1 x} + c_2 e^{\lambda_2 x} + c_3 e^{\lambda_3 x}$$

If the equation turns out to have two complex roots (I won't say if it does), you'll have a solution somewhat different:

$$y(x) = c_1 e^{\lambda_1 x} + c_2 e^{\alpha x} \cos{\beta x} + c_ 3 e^{\alpha x} \sin{\beta x}$$

Where,

$$\lambda_2 = \alpha + \beta i$$
$$\lambda_3 = \alpha - \beta i$$

All of this should be in bold print (well, almost) in any ODE book.

 

[Hurkyl]

how should i solve for (k)

I'd try starting with the rational root theorem, and simply check all the possibilities for a rational solution for k.

(PS wasn't there a 9 in there before?)

 

[graphic7]

(PS wasn't there a 9 in there before?)

There was. Otherwise, I would've mentioned the rational root theorem, as well. Don't you love it when you submit a (generous) post full of tips and someone changes the nature of the problem?

 

[dextercioby]

Here's the solution,courtesy of Maple.

Daniel.