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

[andrew.c]

Homework Statement

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

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

Homework Equations

Knowledge of ODE's
[tex]y = y_{aux}+y_{particular}[/tex]

The Attempt at a Solution

I get the compementary function;
y''+9y = 0

and then using lambda-notation

[tex]\lambda^2 + 9 = 0[/tex]

therefore

[tex]\lambda = \pm 3i[/tex]
(complex roots)

so
[tex]y_{aux} = A cos 3x + B sin 3x[/tex]

but now how do I get [tex]y_{particular}[/tex]??

 

[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.

 

[Mark44]

PI = particular ?

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

 

[andrew.c]

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

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)?

 

[andrew.c]

PI = particular ?

PI = Particular Integral :)

 

[Mark44]

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)?

Not if the complementary solution y_c is c₁cos3x + c₂sin3x, which means that no matter what the values of c₁ and c₂, y_c'' + 9y_c = 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 y_p'' + 9y_p = 36sin(3x).

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