- #1
mitch_1211
- 99
- 1
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 yp = [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 yp = 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
yp = 1/(D2 -2D +3) * xsin(x) + yp = 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
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 yp = [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 yp = 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
yp = 1/(D2 -2D +3) * xsin(x) + yp = 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