Undetermined coefficients

EvLer

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.

Last edited:
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saltydog

Homework Helper
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

Homework Helper
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

$$Y_p(t) = -y_1(t) \int \frac {y_2(t) g(t)}{W(y_1, y_2) (t)} dt ~+~ y_2(t) \int \frac {y_1(t) g(t)}{W(y_1, y_2)(t)}dt$$

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