# Method of Undetermined Coefficients

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1. Feb 27, 2017

### an_single_egg

A question on my homework:

Solve the DE using undetermined coefficients.

y''' - 5y'' + 6y' = 8+2sinx

Ok, supposedly easy. I use a homogenous equation to solve for yc.

y''' -5y'' +6y' = 0
m3-5m2+6m=0
m=0, -1, 6
yc = c1 + c2e-x+c3e6x

Then use a trial solution to find yp, but this is where I get confused...

g(x) = 8+2sinx
yp = A+Bcosx+Csinx

Right...?

yp' = -Bsinx+Ccosx
yp'' = -Bcosx-Csinx
yp''' = Bsinx-Ccosx

Bsinx-Ccosx+5Bcosx+5Csinx-6Bsinx+6Ccosx = 8+2sinx

(-5B+5C) = 2
(5C+5B) = 0

B= -1/5
C=1/5

Here where I run into a problem -- because there is a constant in g(x), but only yp's derivatives are being added together, there is no way to make is equal 8+2sinx, or at least not as far as I can see. Would I just make A=8?

2. Feb 27, 2017

### danyull

The problem comes from the fact that any constant will satisfy the homogenous form of the differential equation (you can see this since you found a constant $c_1$ term in your general solution to the homogenous equation).

The trick is the same as finding second solutions to a homogenous equation with double roots in the auxillary polynomial, which is to add an extra factor of $x$. So try using $Ax+B\cos x+C\sin x$ for your trial solution instead. I expect you'd find that $A$ does indeed equal 8.

3. Feb 28, 2017

### Ray Vickson

An additive constant on the right is always easy to eliminate: just use $z = y' - 8/6$, which obeys the DE
$$z'' -5 z' + 6z = 2 \sin x$$

Last edited: Feb 28, 2017
4. Feb 28, 2017

### ehild

You can solve the equation for y'=z without any difficulty, and get y(x) by integrating.

5. Feb 28, 2017

### an_single_egg

Thanks everyone! I don't think I've learned the second method in my classes yet, but the additive constant thing seemed apparent when I looked at the problem again. :)

6. Mar 2, 2017

### pasmith

It helps to solve the characteristic equation correctly: $m^3 - 5m^2 + 6m = m(m^2 - 5m + 6) = m(m-2)(m-3)$.