Answer to Question #252033 in Differential Equations for Shagor

Question

Solve the method of undermined coefficient (D^2+8D+16)y=8e^-2x ; y(0)= 2 y'(0)= 0

Answer 1

Homogeneous differential equation

$$(D^2+8D+16)y=0$$

Corresponding (auxiliary) equation

$$r^2+8r+16=0$$

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

$$r_1=r_2=-4$$

The general solution of the homogeneous differential equation is

$$y_h=c_1xe^{-4x}+c_2e^{-4x}$$

Find the particular solution of the nonhomogeneous differential equation

$$y_p=Ae^{-2x}$$

$$y_p'=-2Ae^{-2x}$$

$$y_p''=4Ae^{-2x}$$

Substitute

$$4Ae^{-2x}-16Ae^{-2x}+16Ae^{-2x}=8e^{-2x}$$

$$A=2$$

The general solution of the nonhomogeneous differential equation is

$$y=c_1xe^{-4x}+c_2e^{-4x}+2e^{-2x}$$

$y(0)= 2, y'(0)= 0$

$$c_1(0)e^{-4(0)}+c_2e^{-4(0)}+2e^{-2(0)}=2=>c_2=0$$

$$y=c_1xe^{-4x}+2e^{-2x}$$

$$y'=c_1e^{-4x}-4c_1xe^{-4x}-4e^{-2x}$$

$$c_1e^{-4(0)}-4c_1(0)e^{-4(0)}-4e^{-2(0)}=0=>c_1=4$$

The solution of the given initial value problem is

$$y=4xe^{-4x}+2e^{-2x}$$