2nd order differential equations with constant coeff. The Particular integrals.

1. Jul 3, 2008

rock.freak667

For the differential equation

$$\frac{d^2y}{dx^2}+4 \frac{dy}{dx}=sinx$$

One root of the auxiliary equation is '0' meaning the particular integral for the right hand side is x(Asinx+Bcosx). But is there any formal proof for making this claim that for 0 as one root is it is x(Asinx+Bcosx) or 0 were the two roots, the PI would be x^2(Asinx+Bcosx)?

2. Jul 3, 2008

Ben Niehoff

It seems to me that $A \sin x + B \cos x$ will work just fine. You only need the additional factor of $x$ if you have a double root.

To see why it works, simply substitute $y = A \sin x + B \cos x$ into your equation. You should end up with

$$f(A,B) \sin x + g(A,B) \cos x = \sin x$$

To get both sides to be equal for all x, you need to solve two equations in the two unknowns, A and B.

3. Jul 3, 2008

rock.freak667

Ah dumb me I was thinking of the wrong example and made the wrong statement.

But what I really wanted to know is if there is any proof for why a PI should be

x(Acosax+Bsinax) when 'a' is one or both roots of the auxiliary equation (RHS=sine or cosine)
or xe^ax for 'a' as one root and x^2e^ax for 'a' as both roots (RHS=some exp. function)

New example:

$$\frac{d^2y}{dx^2}+4 \frac{dy}{dx}+4y=6e^{-2x}$$

For this example: The PI is of the form $Ax^2e^{-2x}$, but how did we know that we needed to multiply by $x^2$?

4. Jul 3, 2008

Defennder

There's a rule for the choice of polynomial used in the method of undetermined coefficients. I quote this from my notes:

Suppose $r(x)=P_m(x)e^{\mu x}$ is the RHS of the 2nd order linear ODE of polynomial degree m, then the DE has a particular solution of the following form:
$$y= x^k Q_m (x) e^{\mu x}$$
where Qm(x) is an undetermined polynomail with degree m, k is the multiplicity of the root $\mu$ in the characteristic/auxiliary equation $\lambda^2 + a\lambda + b = 0$. If $\mu$ is not a root of the equation, then k=0.

Unfortunately I don't know how to prove that the above rule would always work.

5. Jul 4, 2008

Ben Niehoff

Defennder:

To prove these things, all you need to do is plug the formula in and take the derivatives. Then use induction to prove it for all orders of linear, constant-coefficient ODEs.

6. Jul 4, 2008

Defennder

Differentiating the above twice gives me a very complicated and tedious expression to work with. Ouch.