Obtain the general solution unless otherwise instructed
Given:
$\left({D}^{2}-3D+2\right)y={e}^{2x}{(1+{e}^{2x})}^{-1}$
Formula used:
$y={y}_{h}+{y}_{p}$
$\left({D}^{2}-3D+2\right)y={e}^{2x}{(1+{e}^{2x})}^{-1}$
A linear non-homogeneous ODE with constant coefficient has the form of $\left({a}_{n}{D}^{n}+......+{a}_{1}D+{a}_{0}\right)y=g\left(x\right)$
The general solution to the above expression can be written as follows,
y=y_{h}+y_{p}
Where,
On finding y_{h} by solving (D^{2}-3D+2)y=0
y=c_{1}e^{2x}+c_{2}e^{x}
On finding y_{p} that satisfies (D^{2}-3D+2)y=e^{2x}(1+e^{2x})^{-1}
${y}_{p}={e}^{2x}\left(-\frac{1}{2}\mathrm{ln}\left(1+{e}^{2x}\right)+x\right)-{e}^{x}arc\mathrm{tan}\left({e}^{x}\right)$
Therefore the general solution y=y_{h}+y_{p} is $y={c}_{1}{e}^{2x}+{c}_{2}{e}^{x}+{e}^{2x}\left(-\frac{1}{2}\mathrm{ln}\left(1+{e}^{2x}\right)+x\right)-{e}^{x}arc\mathrm{tan}\left({e}^{x}\right)$