How Do You Apply Undetermined Coefficients to Solve Differential Equations?

[EvLer]

Hello,
I have this DE:

y'' + 2y' - 3y = 8e^x - 12e^3x

when I find homogeneous solution I get

y_h = c₁e^x + 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 = Axe^x,
y₂ = A^3x ?

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.

 

[saltydog]

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:

(D-1)(D-3)y=0

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:

(D-1)(D-3)(D^2+2D-3)y=0

Are you following this?

So the solution to this is:

y_c(x)=C_1e^x+Axe^x+C_3e^{-3x}+Be^{3x}

Now, take:

y_p(x)=Axe^x+Be^{3x} and back-substitute into your original equation, equate coefficients to find A and B.

 

[dextercioby]

Why don't you apply Lagrange's method of variation of constants...?

Daniel.

 

[EvLer]

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.

 

[Corneo]

Lagrange's method of variation of constants is also known as variation of parameters. It is based on the fact that if y_1 and y_2 are solutions to an homogeneous ODE, then so is c_1 y_1 and c_2 y_2 by principle of superposition. But we aim to find the particular solution of the form u(t) y_1(t) and v(t) y_2(t)

The bottom line is that the particular solution is

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