Answer to Question #288033 in Differential Equations for Abhishek

Question

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

Answer 1

Find solution of the associated homogeneous equation:

$$(D^2-3D+2)y=0$$

Characteristic (auxiliary) equation is

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

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

$$r_1=1, r_2=2$$

The general solution of the associated homogeneous equation is

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

Variation of parameters

$$y'=c_1'e^x+c_1e^x+c_2'e^{2x}+2c_2e^{2x}$$

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

$$y'=c_1e^x+2c_2e^{2x}$$

$$y''=c_1'e^x+c_1e^x+2c_2'e^{2x}+4c_2e^{2x}$$

Substitute

$$c_1'e^x+c_1e^x+2c_2'e^{2x}+4c_2e^{2x}$$

$$-3c_1e^x-6c_2e^{2x}+2c_1e^x+2c_2e^{2x}=\dfrac{e^x}{1+e^x}$$

We have

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

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

Then

$$c_1'=-c_2'e^x$$

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

Or

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

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

Integrate

$$c_1=-\int \dfrac{1}{1+e^x}dx=-\int \dfrac{1+e^x-e^x}{1+e^x}dx$$

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

$$c_2=\int \dfrac{e^{-x}}{1+e^x}dx$$

$e^x=u=>du=e^xdx=>dx=\dfrac{du}{u}$

$$c_2=\int \dfrac{e^{-x}}{1+e^x}dx=\int \dfrac{du}{u^2(1+u)}$$

$$\dfrac{1}{u^2(1+u)}=\dfrac{A}{u}+\dfrac{B}{u^2}+\dfrac{C}{1+u}$$

$$=\dfrac{Au(1+u)+B(1+u)+Cu^2}{u^2(1+u)}$$

$$u=0: B=1$$

$$u=-1: C=1$$

$$u=1:2A+2B+C=1=>A=-1$$

$$c_2=\int \dfrac{e^{-x}}{1+e^x}dx=\int \dfrac{du}{u^2(1+u)}$$

$$=-\int \dfrac{du}{u}+\int \dfrac{du}{u^2}+\int \dfrac{du}{1+u}$$

$$=-\ln |u|-\dfrac{1}{u}+\ln(|1+u|)$$

$$=-x-e^{-x}+\ln(1+e^x)$$

The particular solution of the non homogeneous differential equation is

$$y_p=-xe^x+e^x\ln(e^x+1)$$

$$-xe^{2x}-e^{x}+e^{2x}\ln(1+e^x)$$

The general solution of the non homogeneous differential equation is

$$y=c_1e^x+c_2e^{2x}-xe^x+e^x\ln(e^x+1)$$

$$-xe^{2x}-e^{x}+e^{2x}\ln(1+e^x)$$