solution to 2nd order ODE using the D operator method with 2 trig terms on RHS

mitch_1211

Hey,

I have the DE

y'' -2y' + 3y = xsin(x) + 2cosh(2x)

Using the D operator as D = [itex]\frac{dy}{dx}[/itex] this becomes

(D² -2D +3)y = xsin(x) + 2cosh(2x)

so y_p = [itex]\frac{1}{p(D^2)}[/itex] operating on xsin(x) + 2cosh(2x)

(i think)

So i know if this was say [itex]\frac{1}{p(D^2)}[/itex] operating on sin(x)
i.e (D² -2D +3)y = sin(x)
then y_p = 1/(D² -2D +3) * sin(x)

and you would substitute D² = -([itex]\alpha[/itex])² and proceed to solve.(where alpha is the coefficient of x in the argument of sin, here it is 1)

My question is, I have a term on the RHS which is polynomial times trig; xsin(x) and a trig term which can be treated as an exponential; 2cosh(2x)

I know I can split this into
y_p = 1/(D² -2D +3) * xsin(x) + y_p = 1/(D² -2D +3) * 2cosh(2x)

But I'm unsure of the 'rules' to use here, i.e for a single trig term you swapped D²
for -([itex]\alpha[/itex])²
But what do you do for a poly times a trig and for the cosh function?

Any points would be much appreciated, if all else fails I will have to resort to the method of undetermined coefficients