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

  • Thread starter skysurani
  • Start date
  • Tags
    Division
In summary: Since the equation is homogeneous and has constant coefficients, you can just solve its characteristic equation. 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
  • #1
7
0
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)
 
Last edited:
Physics news on Phys.org
  • #2
Is this the equation

[tex] \frac{d^{3}y}{dx^{3}}-3\frac{d^2y}{dx^{2}}+31\frac{dy}{dx}-37y=0 [/tex]

Daniel.
 
  • #3
Since the equation is homogeneous and has constant coefficients, you can just solve its characteristic equation.

You'll have an equation of the form:

[tex]y(x) = c_1 e^{\lambda_1 x} + c_2 e^{\lambda_2 x} + c_3 e^{\lambda_3 x}[/tex]

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

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

Where,

[tex]\lambda_2 = \alpha + \beta i[/tex]
[tex]\lambda_3 = \alpha - \beta i[/tex]

All of this should be in bold print (well, almost) in any ODE book.
 
Last edited:
  • #4
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?)
 
  • #5
Hurkyl said:
(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?
 
  • #6
Here's the solution,courtesy of Maple.

Daniel.
 
Last edited:

1. What is synthetic division?

Synthetic division is a method used to simplify the process of dividing a polynomial by a linear factor. It is commonly used to solve polynomial equations and find their roots.

2. How do I know when to use synthetic division?

Synthetic division is most commonly used when the divisor is in the form of x-c, where c is a constant. This method is also more efficient than long division when the divisor is a quadratic or higher degree polynomial.

3. What is the purpose of using synthetic division to solve an equation?

The purpose of using synthetic division is to find the roots or solutions to a polynomial equation. By setting the equation equal to 0 and using synthetic division, we can test different values for the variable k until we find the value that makes the equation true, thereby solving the equation.

4. Can synthetic division be used for any degree of polynomial?

No, synthetic division can only be used for polynomials with a degree of 2 or higher. For polynomials with a degree of 1, long division or other methods must be used.

5. What are some common mistakes to avoid when using synthetic division?

Some common mistakes to avoid when using synthetic division include incorrectly identifying the coefficients of the polynomial, forgetting to include a placeholder for missing terms, and making calculation errors while performing the division. It is important to double check your work and be careful with the steps involved in synthetic division to avoid these mistakes.

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

Back
Top