Answer to Question #216156 in Differential Equations for Aravind

Question

(D-5D

Answer 1

$$(D^2-5D+6)y=\cos 3x$$

The homogeneous differential equation

$$(D^2-5D+6)y=0$$

The characteristic equation

$$r^2-5r+6=0$$

$$r_1=2,r_2=3$$

The general solution of the homogenepus equation is

$$y_h=c_1e^{2x}+c_2e^{3x}$$

Find the particular solution of the nonhomogeneous differential equation

$$y_p=A\cos3x+B\sin3x$$

$$y_p '=-3A\sin 3x+3B\cos3x$$

$$y_p''=-9A\cos3x-9B\sin3x$$

Substitute

$$-9A\cos3x-9B\sin3x+15A\sin 3x-15B\cos3x$$

$$+6A\cos3x+6B\sin3x=\cos3x$$

$$-3A-15B=1$$ $$15A-3B=0=>B=5A$$

$$-78A=1=>A=-\dfrac{1}{78},B=-\dfrac{5}{78}$$

Then

$$y_p=-\dfrac{1}{78}\cos3x-\dfrac{5}{78}\sin3x$$

The general solution of the given nonhomogenepus equation is

$$y=c_1e^{2x}+c_2e^{3x}-\dfrac{1}{78}\cos3x-\dfrac{5}{78}\sin3x$$