# 2nd order non-homogeneneous ODE - how to find PI

1. Dec 6, 2009

### andrew.c

1. The problem statement, all variables and given/known data

Find the general and, if possible, particular solutions of the following ordinary differential equations:

y''+9y=36sin3x
(hint: modification rule for PI)

2. Relevant equations
Knowledge of ODE's
$$y = y_{aux}+y_{particular}$$

3. The attempt at a solution
I get the compementary function;
y''+9y = 0

and then using lambda-notation

$$\lambda^2 + 9 = 0$$

therefore

$$\lambda = \pm 3i$$
(complex roots)

so
$$y_{aux} = A cos 3x + B sin 3x$$

but now how do I get $$y_{particular}$$??

2. Dec 6, 2009

### rock.freak667

Notice that '3' is a root in the auxiliary equation.

what is the PI for sin3x? Since '3' is a root, you will need to modify the PI by multiplying it by x.

3. Dec 6, 2009

### Staff: Mentor

PI = particular ???

To expand on what rock.freak667 said, your particular solution should be yp = Axcos(3x) + Bxsin(3x).

4. Dec 6, 2009

### andrew.c

I don't really understand what you mean by this..

However, the PI for sin3x is Acos3x + Bsin3x, i think...
so if you have to multiply by x (don't really understand why?) then the PI would be
x(Acos3x + Bsin3x)?

5. Dec 6, 2009

### andrew.c

PI = Particular Integral :)

6. Dec 6, 2009

### Staff: Mentor

Not if the complementary solution yc is c1cos3x + c2sin3x, which means that no matter what the values of c1 and c2, yc'' + 9yc = 0. In other words, there is no way you will end up with 36sin3x on the right side.

Your particular solution (I prefer this term to particular integral) must therefore be Axcos3x + Bxsin3x. You need to find the coefficients A and B so that yp'' + 9yp = 36sin(3x).

Your general solution will be y = yc + yp = c1cos3x + c2sin3x + Axcos3x + Bxsin3x, where you will have determined A and B.