# Trial solution, undetermined coefficients, 2nd order non-homogeneous equation

Tags:
1. Aug 5, 2015

### kostoglotov

1. The problem statement, all variables and given/known data

Write a trial solution for the method of undetermined coefficients. Do not determine the coefficients.

For: $y'' + 2y' + 10y = x^2e^{-x}\cos{3x}$

There's a modification performed and I'm not 100% confident as to why.

2. Relevant equations

3. The attempt at a solution

The complementary equation gives a solution of

$$y_c = e^{-x}\left(c_1\cos{3x} + c_2\sin{3x}\right)$$

Now, before considering a modification to the particular solution $y_p$, let's consider the standard trial solution by the method of undetermined coefficients. I would say it's

$$y_p = e^{-x}\left[(Ax^2+Bx+C)\cos{3x} + (Dx^2+Ex+F)\sin{3x}\right]$$

Now I know that the modification is performed so that "no term in the particular solution can be a solution for the complementary equation"...I do know basically what this means...and I would have thought that for the particular trial solution above, that neither term would suitably solve the complementary equation. Because the particular solution has a polynomial of degree 2 and the complementary solution has a polynomial of degree 0.

So I've thought, maybe we could solve the complementary equation with the trial solution $y_p$ if we set A,B and/or D,E to = 0,0, so that the polynomials in the trial solutions were reduced from 2nd degree to 0 degree.

Is this why we need to modify $y_p$ by multiply it by x? Because we can set A,B,D,E to zero in order to solve for the complementary equation? Thus, it won't really matter what degree a polynomial is in the solutions as far as deciding to modify is concerned?

2. Aug 6, 2015

### ehild

As e-xcos(3x) is solution of the homogeneous (complementary) equation, you have to modify the particular solution by multiplying it with x.

Last edited: Aug 6, 2015
3. Aug 6, 2015

### kostoglotov

Yes I understand that, but in my particular solution I've already multiplied it by a polynomial of 2nd degree...why does it need to be brought up to a polynomial of 3rd degree?

4. Aug 6, 2015

### ehild

You had to multiply with second order polynomials as there is the factor x2 in the inhomogeneous term. One more x to multiply with is because e-xcos(3x) is solution of the homogeneous equation. Try to find a solution with your method; you won't be able.

Last edited: Aug 6, 2015
5. Aug 7, 2015

### vela

Staff Emeritus
Denote the derivative operator by D. The operator D+r acting on a term of the form $x^p e^{-rx}$ yields
$$(D+r) (x^p e^{-rx}) = (p x^{p-1} e^{-rx} - r x^p e^{-rx}) + r x^p e^{-rx} = p x^{p-1} e^{-rx}.$$ The effect of the operator is basically to reduce the power of $x$ by 1 if $p\ne 0$. If p=0, we have $(D+r)e^{-rx} = 0$. In other words, $e^{-rx}$ is a solution of the homogeneous differential equation y' + ry = 0.

Now consider the differential equation $(D+r)y = x^n e^{-rx}$. It has a homogeneous solution $y_h = e^{-rx}$. From the result above, you should be able to see that in order to get a term like $x^n e^{-rx}$, $y$ must contain a term of the form $x^{n+1} e^{-rx}$. It doesn't matter that $x^n e^{-rx}$, …, and $x e^{-rx}$ are independent from $y_h$. When D+r acts on those terms, you end up with $x^{n-1}e^{-rx}$, …, and $e^{-rx}$, and there's no linear combination of them that is equal to $x^n e^{-rx}$.

Last edited: Aug 7, 2015