Answer to Question #225224 in Differential Equations for Asif

Question

Solve the differential equation 2y^'''+ y^''- 8y^' - 4y = 7 - 2e^-x - cos² (2x).

Answer 1

The homogeneous equation

$$2y'''+y''-8y'-4y=0$$

The characteristic equation

$$2r^3+r^2-8r-4=0$$

$$r^2(2r+1)-4(2r+1)=0$$

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

$$r_1=-\dfrac{1}{2}, r_2=-2, r_3=2$$

The general solution of the homogeneous differential equation is

$$y_h=C_1e^{-x/2}+C_2e^{-2x}+C_3e^{2x}$$

$$7-2e^{-x}-\cos^2(2x)=7-2e^{-x}-\dfrac{1}{2}-\dfrac{1}{2}\cos(4x)$$

$$=\dfrac{13}{2}-2e^{-x}-\dfrac{1}{2}\cos(4x)$$

Find the particular solution of the nonhomogeneous differential equation

$$y_p=A+Be^{-x}+C\cos(4x)+D\sin(4x)$$

$$y_p'=-Be^{-x}-4C\sin(4x)+4D\cos(4x)$$

$$y_p''=Be^{-x}-16C\cos(4x)-16D\sin(4x)$$

$$y_p'''=-Be^{-x}+64C\sin(4x)-64D\cos(4x)$$

Substitute

$$-2Be^{-x}+128C\sin(4x)-128D\cos(4x)$$

$$+Be^{-x}-16C\cos(4x)-16D\sin(4x)$$

$$+8Be^{-x}+32C\sin(4x)-32D\cos(4x)$$

$$-4A-4Be^{-x}-4C\cos(4x)-4D\sin(4x)$$

$$=\dfrac{13}{2}-2e^{-x}-\dfrac{1}{2}\cos(4x)$$

$$A=-\dfrac{13}{8}$$

$$B=-\dfrac{2}{3}$$

$$D=8C$$

$$C=\dfrac{1}{2600}$$

$$D=\dfrac{1}{325}$$

Then

$$y_p=-\dfrac{13}{8}-\dfrac{2}{3}e^{-x}+\dfrac{1}{2600}\cos(4x)+\dfrac{1}{325}\sin(4x)$$

The general solution of the nonhomogeneous differential equation is

$$y=C_1e^{-x/2}+C_2e^{-2x}+C_3e^{2x}$$

$$-\dfrac{13}{8}-\dfrac{2}{3}e^{-x}+\dfrac{1}{2600}\cos(4x)+\dfrac{1}{325}\sin(4x)$$