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?

  • $\begingroup$ There must be a typo, neither $e^x$ nor $e^{2x}$ is a solution of the DE. Did you mean $y''-3y'+2y$? $\endgroup$ – André Nicolas Feb 23 '12 at 20:48
  • $\begingroup$ @AndréNicolas It seems like it. $\endgroup$ – Pedro Tamaroff Feb 23 '12 at 20:51
  • $\begingroup$ @AndréNicolas: sorry I was too uncareful, fixed that. Thanks, now back to terms. $\endgroup$ – hhh Feb 23 '12 at 20:55
  • $\begingroup$ 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. $\endgroup$ – André Nicolas Feb 23 '12 at 21:01
  • 1
    $\begingroup$ 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. $\endgroup$ – 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 $$y=y_{hom}+y_p=C_1e^{2x} + C_2e^x + \frac{2-w^2}{w^4+5w^2+4}\cos(wx) + \frac{-3w}{w^4+5 w^2+4}\sin(wx)$$

  • 1
    $\begingroup$ With the tagged equation by (1), is it always so that the particular solution is of that form or just with trigonometric RHS? $\endgroup$ – hhh Feb 24 '12 at 0:45
  • $\begingroup$ 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. $\endgroup$ – Andrew Parker Feb 24 '12 at 1:34
  • $\begingroup$ @Peter - if it's worth an edit, it's worth an upvote? $\endgroup$ – Andrew Parker Feb 25 '12 at 6:43
  • $\begingroup$ @AndrewParker Hmm, most probably. I just added the slashes. $\endgroup$ – Pedro Tamaroff Feb 25 '12 at 14:21
  • $\begingroup$ @Peter I always forget about that with the trig functions... thanks. $\endgroup$ – Andrew Parker Feb 25 '12 at 16:17

I just wanted to answer the original question and point something out. In every class I've ever had, the complementary solution is the solution to the "homogeneous equation" and is often (if incorrectly) referred to as the homogeneous solution. So in this case I believe you are correct. The original equation is not itself homogeneous; however, the "homogeneous solution" is the solution for the original equation without a driving function (RHS = 0). And you've put that as your answer.

**Another point, as homogeneous solution and complementary solution are often used interchangeably, "particular solution" and "complimentary solution" are definitely not used interchangeably. In several answers and comments, people sound is if they refer to the same thing when they do not. For any linear ordinary differential equation, the general solution (for all t for the original equation) can be represented as the sum of the complementary solution and the particular solution. Vg(t)=Vp(t)+Vc(t)

In electrical engineering context, the complementary and particular solutions have their own names (because we always rename things), the particular solution is usually called the "steady-state response" and the complementary solution is called the "transient response".


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.