Question

Решить уравнение:
y'' + y'=$frac{1}{1+e^{x} }$

Answer 1

$y''+y'=dfrac{1}{1+e^x}\ e^xy''+e^xy'=dfrac{e^x}{1+e^x}\ (e^xy')'_x=dfrac{e^x}{1+e^x}\ e^xy'=intdfrac{e^x}{1+e^x}dx=intdfrac{1}{1+e^x}d(e^x+1)=ln(e^x+1)+C_1\ y'=dfrac{ln(e^x+1)}{e^x}+dfrac{C_1}{e^x}$

$y=intdfrac{ln(e^x+1)}{e^x}dx+intdfrac{C_1}{e^x}dx=(*)\ intdfrac{ln(e^x+1)}{e^x}dx=int ln(1-dfrac{1}{-frac{1}{e^x}}) d(-dfrac{1}{e^x})=int ln(1-dfrac{1}{t}) dt=[u=ln(1-dfrac{1}{t}),;du=dfrac{dfrac{1}{t^2}}{(1-dfrac{1}{t})}dt,dv=1,v=t]=t*ln|1-dfrac{1}{t}|-int dfrac{1}{t-1}dt=t*ln|1-dfrac{1}{t}|-ln|t-1|+C_2=-e^{-x}ln(1+e^{x})-ln(e^{-x}+1)+C_2\ (*)=-e^{-x}ln(1+e^{x})-ln(e^{-x}+1)+C_3e^{-x}+C_4$