Undetermined coefficients

In summary, the conversation is about solving a differential equation with a particular solution using the method of undetermined coefficients. One person suggests using Lagrange's method of variation of constants, while another person suggests solving the homogeneous equation first and then using the principle of superposition to find the particular solution. The conversation ends with the mention of the particular solution formula.
  • #1
EvLer
458
0
Hello,
I have this DE:

y'' + 2y' - 3y = 8ex - 12e3x

when I find homogeneous solution I get

yh = c1ex + c2e-3x;

so now to find the particular solution by method of undetermined coefficients, do I set y to smth like this:

y = y1 + y2

where
y1 = Axex,
y2 = A3x ?

since one of the solutions to auxiliary equation appears on the RHS of the DE and the other does not?
I don't need the full solution, just confirmation/correction of this part.

Thanks much!

EDIT: if I take the fact that if y1 + y2 is a solution, then y1 is a solution and y2 is a solution. I guess I answered my own question. :frown:
 
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  • #2
Well EvLer, it's nice if you first recognize that the RHS is a particular solution to a homogeneous equation with roots 1, 3 so that equation would be:

[tex](D-1)(D-3)y=0[/tex]

You familiar with those differential operations right?

Thus, applying that operator to both sides of your equation will make the RHS 0 right (it's a solution to that homogeneous operator which is set to zero). So applying it we get:

[tex](D-1)(D-3)(D^2+2D-3)y=0[/tex]

Are you following this?

So the solution to this is:

[tex]y_c(x)=C_1e^x+Axe^x+C_3e^{-3x}+Be^{3x}[/tex]

Now, take:

[tex]y_p(x)=Axe^x+Be^{3x}[/tex] and back-substitute into your original equation, equate coefficients to find A and B.
 
  • #3
Why don't you apply Lagrange's method of variation of constants...?

Daniel.
 
  • #4
Yeah, thanks, I got it 5 minutes after posting.

@saltydog: it's the same technique, only we follow it more step-by-step, where we solve homogeneous equation first and then based on the solution of non-hom. and hom. eq. we pick appropriate form of particular solution, and then do differentiation and plug it all in.

ARRRGGGH @ DE!

ps: don't know what "Lagrange's method of variation of constants" is but thanks, i'll look that up.
 
  • #5
Lagrange's method of variation of constants is also known as variation of parameters. It is based on the fact that if [itex]y_1[/itex] and [itex]y_2[/itex] are solutions to an homogeneous ODE, then so is [itex]c_1 y_1[/itex] and [itex]c_2 y_2[/itex] by principle of superposition. But we aim to find the particular solution of the form [itex]u(t) y_1(t)[/itex] and [itex]v(t) y_2(t)[/itex]

The bottom line is that the particular solution is

[tex]Y_p(t) = -y_1(t) \int \frac {y_2(t) g(t)}{W(y_1, y_2) (t)} dt ~+~ y_2(t) \int \frac {y_1(t) g(t)}{W(y_1, y_2)(t)}dt[/tex]
 

1. What is the concept of undetermined coefficients?

Undetermined coefficients is a method used in mathematics to solve linear differential equations with non-constant coefficients. It involves finding a particular solution by guessing a solution based on the form of the non-homogeneous term in the equation.

2. How does the method of undetermined coefficients work?

The method of undetermined coefficients involves guessing a particular solution to a non-homogeneous linear differential equation based on the form of the non-homogeneous term. This solution is then substituted into the original equation to determine the coefficients, which are then used to find the general solution.

3. What types of equations can be solved using undetermined coefficients?

Undetermined coefficients can be used to solve linear differential equations with non-constant coefficients, as long as the non-homogeneous term is a polynomial, exponential, sine, cosine, or a combination of these functions.

4. Are there any limitations to the method of undetermined coefficients?

Yes, the method of undetermined coefficients can only be used to find a particular solution when the non-homogeneous term is a polynomial, exponential, sine, cosine, or a combination of these functions. It cannot be used to solve equations with non-linear or non-constant coefficients.

5. How is the particular solution found using undetermined coefficients different from the general solution?

The particular solution found using undetermined coefficients is a specific solution that satisfies the non-homogeneous term in the equation, while the general solution includes all possible solutions, including the particular solution, to the homogeneous and non-homogeneous terms.

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