# Undetermined Coefficients

1. Mar 18, 2009

### kuahji

Undetermined Coefficients to solve.

y'''-6y''=3-cosx

So I set everything up, get my system & I get an answer of

y=c1+c2x+c3e^(6x)-6/37cosx+1/37sinx

the book has y=c1+c2x+c3e^(6x)-6/37cosx+1/37sinx-1/4x^2.

So I'm not sure where the book gets the last term, any ideas?

2. Mar 18, 2009

### gabbagabbahey

The book's answer is correct, but I have no idea where you are going wrong since you haven't shown your work...

3. Mar 18, 2009

### kuahji

Ok, I have yp= A + B cosx + c sinx
y'p=-Bsinx+C cosx
y''p=-B cosx - C sinx
y'''P=Bsinx - C cosx

After I have plugged these back into the LHS, I get a system of
B+6C=0
6B-C=-1

These give me my coefficients -6/37cosx & 1/37sinx.

4. Mar 18, 2009

### kuahji

Oh & yh=r^2(r-6), which is where I get c1+c2x+c3e^6c.

5. Mar 18, 2009

### gabbagabbahey

Well that's your problem right there, the constant term 'A' is not linearly independent to your homogeneous solution, which also contains a constant term.... to account for the constant term (3) on the RHS of your DE, you need to add a term of the form Ax^r where r is the smallest non-negative integer such that no terms of the same order are present in your homogeneous solution....

6. Mar 18, 2009

### kuahji

Thanks. It works out nicely now.