# Thread: Solving for A,B w/ Method of Undetermined Coefficients

1. ## Solving for A,B w/ Method of Undetermined Coefficients

In trying to find the particular solution to a non-homogeneous third-order ODE, I used the method of undetermined coefficients but it seems to be the method of undeterminable coefficients.

I have

y^(3) - y^(1) = 2sint

So I assumed

Y(t) = Acost + Bsint

Differentiating and plugging in, I get

(Asint - Bcost)^3 + Asint - Bcost = 2sint

In solving these problems before, I would follow the idea, "There is a 0cost on the right, so the collected coefficients of the cost on the left must be zero." Applying this to the (sint)^3, (sint)^2(cost), (sint)(cost)^2, (cost)^3, sint, and cost, I would think that

A^3 = 0
-(A^2)B = 0
A(B^2) = 0
-B^3 = 0
A = 2
-B = 0

However, A cannot both be two and its cube be zero. Also, the back of the book gives Y(t) as cost, for which, if Y(t) = Acost + Bsint, A would be one.

I know I am solving for the coefficients incorrectly, but I don't know why my method is wrong here (since I know it has worked in other situations). How can I solve for these coefficients? Is my assumption of what Y(t) is wrong? Should it be something other than Acost + Bsint?

Thank you!

2. Originally Posted by Oijl
In trying to find the particular solution to a non-homogeneous third-order ODE, I used the method of undetermined coefficients but it seems to be the method of undeterminable coefficients.

I have

y^(3) - y^(1) = 2sint

So I assumed

Y(t) = Acost + Bsint

Differentiating and plugging in, I get

(Asint - Bcost)^3 + Asint - Bcost = 2sint

In solving these problems before, I would follow the idea, "There is a 0cost on the right, so the collected coefficients of the cost on the left must be zero." Applying this to the (sint)^3, (sint)^2(cost), (sint)(cost)^2, (cost)^3, sint, and cost, I would think that

A^3 = 0
-(A^2)B = 0
A(B^2) = 0
-B^3 = 0
A = 2
-B = 0

However, A cannot both be two and its cube be zero. Also, the back of the book gives Y(t) as cost, for which, if Y(t) = Acost + Bsint, A would be one.

I know I am solving for the coefficients incorrectly, but I don't know why my method is wrong here (since I know it has worked in other situations). How can I solve for these coefficients? Is my assumption of what Y(t) is wrong? Should it be something other than Acost + Bsint?

Thank you!
I guess I 'm a little confused over your notation...

$\displaystyle \frac{d^3y}{dt^3}-\frac{dy}{dt}=2\sin(t)$

Is this is the case your auxillary equation is

$\displaystyle m^3-m=0 \iff m(m-1)(m+1)$

so $\displaystyle y_c=c_1+c_2e^{t}+c_3e^{-t}$

Now $\displaystyle y_p=A\sin(t)+B\cos(t)$\

$\displaystyle \frac{dy_p}{dt}=A\cos(t)-B\sin(t)$

$\displaystyle \frac{d^3y_p}{dt^3}=-A\cos(t)+B\sin(t)$

Plugging into the original we get

$\displaystyle -A\cos(t)+B\sin(t)-(A\cos(t)-B\sin(t))=2\sin(t)$

$\displaystyle -2A\cos(t)+2B\sin(t)=2\sin(t)$

This tells us that A=0 and B=1

So $\displaystyle y_p=\cos(t)$

3. Doh! Of course, thanks, just needed another pair of eyes. I had plugged the derivatives of the particular solution into the auxiliary equation, somehow. I need better bookkeeping while doing problems, I guess. Thanks.