Answer to Question #139805 in Differential Equations for Nikhil Singh

Question

Solve, by using the method of variation of parameter, the following differential equation
(D^2-3D+2)y = 1/(1+e^-x)

Answer 1

$\displaystyle (D^2 - 3D + 2)y = \frac{1}{1 + e^{-x}}\\ \textsf{The auxiliary equation is}\\ m^2 - 3m + 2 = 0\\ (m - 1)(m - 2) = 0\\ m = 1, 2\\ y = y_p + y_c\\ y_c \, \textsf{is the complementary factor}\\ y_p \, \textsf{is the particular Integral}\\ y_c = Ae^x + Be^{2x}\\ y_p = V_1(x)e^{2x} + V_2(x)e^{x}\\ W(x) \, \textsf{is the Wronskian of the solution}\\ W(x) = e^{\int 3 \, \mathrm{d}x}= e^{3x}\\ \begin{aligned} V_1(x) &= -\int \frac{e^x}{e^{3x}(1 + e^{-x})}\, \mathrm{d}x\\ & = -\int \frac{e^{-2x}}{1 + e^{-x}}\, \mathrm{d}x\\ &= -\int \frac{e^{-x}\cdot e^{-x}}{1 + e^{-x}}\, \mathrm{d}x\\ &= \int \frac{e^{-x}}{1 + e^{-x}}\, \mathrm{d}(e^{-x})\\ &= \int \left(1 - \frac{1}{1 + e^{-x}}\right) \mathrm{d}(e^{-x})\\ &= e^{-x} - \ln(1 + e^{-x}) + C \end{aligned}\\ \begin{aligned} V_2(x) &= \int \frac{e^{2x}}{e^{3x}(1 + e^{-x})}\, \mathrm{d}x\\ & = \int \frac{e^{-x}}{1 + e^{-x}}\, \mathrm{d}x\\ &= -\int \frac{1}{1 + e^{-x}}\, \mathrm{d}(e^{-x})\\ &= -\ln(1 + e^{-x}) + C \end{aligned}\\ \textsf{The constants can be ignored, since we are}\\\textsf{trying to discover non-constant solution to the DE}\\ \begin{aligned} y &= Ae^{x} + Be^{2x} -e^{x}\ln(1 + e^{-x}) \\&+ e^{2x}(e^{-x} - \ln(1 + e^{-x})) \end{aligned}\\ y = Ae^{x} + Be^{2x} -e^{x}\ln(1 + e^{-x}) + e^{x} - e^{2x}\ln(1 + e^{-x})\\ \therefore y = Ae^{x} + Be^{2x} - (e^{x}+e^{2x})\ln(1 + e^{-x}) + e^x$