Linear Second Order ODE/y'' - 2 y' + y = 1 over 1 + e^x

Theorem

The second order ODE:

$(1): \quad y - 2 y' + y = \dfrac 1 {1 + e^x}$

has a particular solution:

$y = 1 + e^x \ds \int \map \ln {1 + e^{-x} } \rd x$

Proof

\(\ds \paren {D^2 - 2 D + 1} y\)

\(=\)

\(\ds \dfrac 1 {1 + e^x}\)

expressing $(1)$ in a different form

\(\ds \leadsto \ \ \)

\(\ds \paren {D^2 - 1}^2 y\)

\(=\)

\(\ds \dfrac 1 {1 + e^x}\)

and in a different form again

\(\ds \leadsto \ \ \)

\(\ds \paren {D - 1} z\)

\(=\)

\(\ds \dfrac 1 {1 + e^x}\)

putting $\paren {D - 1} y = z$

\(\ds \leadsto \ \ \)

\(\ds \map D {z e^{-x} }\)

\(=\)

\(\ds \dfrac {e^{-x} } {1 + e^x}\)

Solution to Linear First Order ODE with Constant Coefficients: integrating factor $e^x$

\(\ds \)

\(=\)

\(\ds \dfrac {e^{-2 x} } {1 + e^{-x} }\)

multiplying top and bottomof right hand side by $e^{-x}$

\(\ds \leadsto \ \ \)

\(\ds z e^{-x}\)

\(=\)

\(\ds \int \dfrac {e^{-2 x} } {1 + e^{-x} } \rd x\)

integrating both sides with respect to $x$

\(\ds \)

\(=\)

\(\ds -\int \dfrac u {1 + u} \rd u\)

Integration by Substitution: setting $e^{-x} = u$

\(\ds \)

\(=\)

\(\ds -\int 1 - \dfrac 1 {1 + u} \rd u\)

\(\ds \)

\(=\)

\(\ds -u + \map \ln {1 + u}\)

Primitive of Constant, Primitive of Reciprocal

\(\ds \)

\(=\)

\(\ds -e^{-x} + \map \ln {1 + e^{-x} }\)

substituting back for $u$

Then we need to solve:

\(\ds \paren {D - 1}y\)

\(=\)

\(\ds z\)

\(\ds \leadsto \ \ \)

\(\ds \map D {y e^{-x} }\)

\(=\)

\(\ds z e^{-x}\)

\(\ds \leadsto \ \ \)

\(\ds y e^{-x}\)

\(=\)

\(\ds e^{-x} + \int \map \ln {1 + e^{-x} } \rd x\)

\(\ds \leadsto \ \ \)

\(\ds y\)

\(=\)

\(\ds 1 + e^x \int \map \ln {1 + e^{-x} } \rd x\)

and so on.

$\blacksquare$

Sources

• 1958: G.E.H. Reuter: Elementary Differential Equations & Operators ... (previous) ... (next): Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 2$. The second order equation: $\S 2.8$ Recapitulation: Example $10$