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

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

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