Answer to Question #210825 in Differential Equations for kanto@49

Question

(D2+3D+2)y=cos2x

Answer 1

$$y''+3y'+2y=\cos2x$$

Write the related homogeneous or complementary equation:

$$y''+3y'+2y=0$$

The general solution of a nonhomogeneous equation is the sum of the general solution $y_h(x)$ of the related homogeneous equation and a particular solution $y_p(x)$ of the nonhomogeneous equation:

$$y(x)=y_h(x)+y_p(x)$$

Consider a homogeneous equation 

$$y''+3y'+2y=0$$

Write the characteristic (auxiliary) equation:

$$r^2+3r+2=0$$

$$(r+2)(r+1)=0$$

$$r_1=-2, r_2=-1$$

The general solution of the homogeneous equation is

$$y_h(x)=C_1e^{-2x}+C_2e^{-x}$$

Let

$$y_p=A\sin 2x+B\cos 2x$$

Then

$$y_p'=2A\cos 2x-2B\sin 2x$$

$$y_p''=-4A\sin2x-4B\cos 2x$$

$$+x(-A\sin x-B\cos x)$$

Substitute

$$-4A\sin2x-4B\cos 2x+6A\cos 2x-6B\sin 2x$$

$$+2A\sin 2x+2B\cos 2x=\cos 2x$$

$$-2A-6B=0$$

$$6A-2B=1$$

$$A=\dfrac{3}{20}, B=-\dfrac{1}{20}$$

The general solution of a second order nonhomogeneous differential equation be

$$y(x)=C_1e^{-2x}+C_2e^{-x}+\dfrac{3}{20}\sin 2x-\dfrac{1}{20}\cos 2x$$