Question

y''+3y'+2y= 1/(1+e^x)

Answer 1

Related homogeneous differential equation

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

The roots of the characteristic equation are

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

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

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

The general solution of the homogeneous differential equation is

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

Use the method of variation of parameters

$$c_1'e^{-2x}+c_2'e^{-x}=0$$

$$c_1'(e^{-2x})'+c_2'(e^{-x})'=\dfrac{1}{1+e^x}$$

Then

$$c_2'e^{-x}=-c_1'e^{-2x}$$

$$-2c_1'e^{-2x}+c_1'e^{-2x}=\dfrac{1}{1+e^x}$$

$$c_1'=-\dfrac{e^{2x}}{1+e^x}$$

$$c_1=-\int\dfrac{e^{2x}}{1+e^x}dx=\ln(e^x+1)-e^x+C_1$$

$$c_2'=\dfrac{e^{x}}{1+e^x}$$

$$c_2=\int\dfrac{e^{x}}{1+e^x}dx=\ln(e^x+1)+C_2$$

The general solution of the homogeneous differential equation is

$$y=e^{-2x}\ln(e^x+1)+C_1e^{-2x}+e^{-x}\ln(e^x+1)+C_3e^{-x}$$