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

[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)?

[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.

[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?

[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.

[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.

[Defennder]

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