# Complementary Solution = Homogenous solution?

I have calculated solutions to homogenous equations but is the complementary solution mentioned here the same as the homogenous solution?

Let's take example $y''-3y'+2y=\cos(wx)$ and now the homogenous solution is

$$y_{hom}=C_{1}e^{2x}+C_{2}e^{x}_{|\text{Characteristic eq. =}(r-1)(r-2)}$$

which can be showed with Wronk's determinant to be valid (cannot yet understand it but go on). Now to find out the general solution there are multiple ways apparently:

1. Method of Undetermined Coefficient
2. Variation of constant

I have not practised them yet enough (because cannot understand the terms yet) so cannot ask much about them but I am trying to, could someone help me with the terminology here about the complementary solution?

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There must be a typo, neither $e^x$ nor $e^{2x}$ is a solution of the DE. Did you mean $y''-3y'+2y$? –  André Nicolas Feb 23 '12 at 20:48
@AndréNicolas It seems like it. –  Pedro Tamaroff Feb 23 '12 at 20:51
@AndréNicolas: sorry I was too uncareful, fixed that. Thanks, now back to terms. –  hhh Feb 23 '12 at 20:55
The usual term is "particular solution." Once you find a particular solution $f_p(x)$, the general solution is $f_p(x)$ plus the general solution of the homogeneous equation. –  André Nicolas Feb 23 '12 at 21:01
As remarked, that is the complementary solution or a particular solution. It is best not to call it a "homogeneous solution" since in fact it is not homogeneous. –  GEdgar Feb 24 '12 at 1:22

As Andre mentioned in his comment, the more common terminology is "particular solution".

Your homogeneous solution $y_{hom} = C_1e^{2x} + C_2e^x$ a solution, not to the original equation, but to the homogeneous equation $y'' - 3y' + 2y = 0$, regardless of the constant parameters $C_1$ and $C_2$.

To find the particular (complementary) solution, we must consider solutions of the form $$y_p=A\cos(wx) + B\sin(wx)\tag{1}$$ After finding $$y_p'=wB\cos(wx) - wA\sin(wx)$$ and $$y_p''=-w^2A\cos(wx) - w^2B\sin(wx)$$ we substitute them into the original equation to get:

$-w^2A\cos(wx) - w^2Bsin(wx) - 3wB \cos(wx) + 3wA\sin(wx) + 2A\cos(wx) + 2Bs\in(wx) = \cos(wx)$

Because $\sin(wx)$ and $\cos(wx)$ are linearly independent, we know that the coefficients of $\cos(wx)$ on the LHS must equal the coefficient of $\cos(wx)$ on the RHS, and similarly for $\sin(wx)$ which gives us the system of equations:

$$\begin {cases} (-w^2 + 2)A + (-3w)B = 1 \\ (3w)A + (-w^2 + 2)B = 0 \end{cases}$$

Solving this system of equations gives us specific values for the coefficients $A$ and $B$:

$$A = \frac{2-w^2}{w^4+5w^2+4}, B = \frac{-3w}{w^4+5 w^2+4}$$

(provided that the denominator of those fractions is non-zero of course - also notice that that is equivalent to the condition that the determinant of the coefficient matrix for $A$ and $B$, in the system of equations given above, is non-zero.)

which now provides us with a particular (complimentary) solution to the original differential equation:

$$y_p=\frac{2-w^2}{w^4+5w^2+4}cos(wx) + \frac{-3w}{w^4+5 w^2+4}sin(wx)$$

And because, in general, (f + g)' = f' + g', we can see that

$$(y_{hom} + y_p)'' - 3(y_{hom} + y_p)' + 2(y_{hom} + y_p) = (y_{hom}'' - 3y_{hom}' + 2y_{hom}) + (y_p'' -3y_p' +2y_p)$$ and then $$(y_{hom}'' - 3y_{hom}' + 2y_{hom}) + (y_p'' -3y_p' +2y_p) = 0 + cos(wx) = cos(wx)$$ thus the general solution will be $$y=y_{hom}+y_p=C_1e^{2x} + C_2e^x + \frac{2-w^2}{w^4+5w^2+4}\cos(wx) + \frac{-3w}{w^4+5 w^2+4}\sin(wx)$$

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With the tagged equation by (1), is it always so that the particular solution is of that form or just with trigonometric RHS? –  hhh Feb 24 '12 at 0:45
Good question. (1) comes from the fact that our non-homogeneous function, $cos(wx)$, has a finite set of linearly independent derivatives - consisting of just $cos(wx)$ and $sin(wx)$. Caveat: Special steps must be taken when the set of linearly independent derivatives intersect with the homogeneous solution. –  Andrew Parker Feb 24 '12 at 1:34
@Peter - if it's worth an edit, it's worth an upvote? –  Andrew Parker Feb 25 '12 at 6:43
@AndrewParker Hmm, most probably. I just added the slashes. –  Pedro Tamaroff Feb 25 '12 at 14:21
@Peter I always forget about that with the trig functions... thanks. –  Andrew Parker Feb 25 '12 at 16:17