Undetermined Coefficients to solve

394
1
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?
 

gabbagabbahey

Homework Helper
Gold Member
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The book's answer is correct, but I have no idea where you are going wrong since you haven't shown your work...
 
394
1
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.
 
394
1
Oh & yh=r^2(r-6), which is where I get c1+c2x+c3e^6c.
 

gabbagabbahey

Homework Helper
Gold Member
5,002
6
Ok, I have yp= A + B cosx + c sinx
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....

where I get c1+c2x+c3e^6x
 
394
1
Thanks. It works out nicely now.
 

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