# Need help understanding an aspect of undetermined coeff's

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1. Jan 19, 2016

### kostoglotov

I am comfortable with the need to multiply the polynomial in front of $e^{2x}$ by $x^3$, that makes perfect sense in terms of what the text has already said about how no term in the particular solution should duplicate a term in the complementary solution, and that itself is logical in terms of the linear independence of the solutions.

What I don't fully understand is the need to multiply the polynomials in front of the trig functions by $x$.

$(D+Ex)\cos{3x} + (F+Gx)\sin{3x}$ already has non-duplicated terms in it, ie, $Ex\cos{3x}$ and $Gx\sin{3x}$.

Why wouldn't it be sufficient to simply use $(D+Ex)\cos{3x} + (F+Gx)\sin{3x}$ and assume that D and F will be zero or else absorbed by linearity into the undetermined constant coefficients of the corresponding trig functions in the complementary solution?

I have of course tried this, and it turns out that indeed the $x\cos{3x}$ and $x\sin{3x}$ terms cancel out when substituted back into the original differential equation, which is

I've subbed the trial solution $(D+Ex)\cos{3x} + (F+Gx)\sin{3x}$ into $(D^2+9)y$ (the factor in the diff eq that produces the trig components of the RHS), and I can see how in this instance the $x\cos{3x}$ and $x\sin{3x}$ terms cancel out, leaving no $x\sin{3x}$ term.

I have come to agree with the insufficiency of $(D+Ex)\cos{3x} + (F+Gx)\sin{3x}$ in this particular case, but I don't know why or how this carries over to the general case.

Why wouldn't $(D+Ex)\cos{3x} + (F+Gx)\sin{3x}$ generally be okay in the particular solution when no $x\sin{3x}$ or $x\cos{3x}$ terms appear in the complementary solution?

2. Jan 19, 2016

### HallsofIvy

Because sin(3x) and cos(3x) are part of the solution to the homogeneous equation, "Ax sin(3x)+ Bx cos(3x)" would give a solution to an equation with non-homogeneous part "sin(3x)+ cos(3x)". In order to get a solution with non-homogeneous part "x sin(3x)+ x cos(3x)" you have to multiply by x yet again, using $x^2 sin(3x)+ x^2 cos(3x)$. In general, if you have a non-homogenous part $$x^n sin(3x)$$ and/or $$x^n cos(3x)$$ you need to use $$Ax^{n+1} sin(3x)+ Bx^{n+1} cos(3x)$$.

Last edited by a moderator: Jan 19, 2016