Use the annihilator method and find the general solution of $y''-3y'+2y=4\sin^3(3x)$
Use the annihilator method and find the general solution of
$$y''-3y'+2y=4\sin^3(3x)$$ $$m^2-3m+2=0$$ $$(m-1)(m-2)=0$$ $$m=1,m=2$$ $$y_c=c_1e^x+c_2e^{2x}$$
Wich is the annihilator of $4\sin^3(3x)$?
Solution 1:
Hint: Try using
$$4\sin^3(3x) = 3 \sin (3 x)-\sin (9 x)$$
Using the Annihilator Method, this will lead to the result $y(x) = y_h(x)+ y_p(x)$
$$y(x) = c_1 e^x+c_2 e^{2 x}+\dfrac{27}{130} \cos (3 x)-\dfrac{21}{130} \sin (3 x)-\dfrac{27 }{6970}\cos (9 x)+\dfrac{79 }{6970}\sin (9 x)$$