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I have the DE

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

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

(D^{2}-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 (D^{2}-2D +3)y = sin(x)

then y_{p}= 1/(D^{2}-2D +3) * sin(x)

and you would substitute D^{2}= -([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/(D^{2}-2D +3) * xsin(x) + y_{p}= 1/(D^{2}-2D +3) * 2cosh(2x)

But I'm unsure of the 'rules' to use here, i.e for a single trig term you swapped D^{2}

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

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

Can you offer guidance or do you also need help?

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