Question

# (D² – 3D + 2)y = e2×(1 + e2x)¯ - ...

Obtain the general solution unless otherwise instructed

Transcribed: (D² – 3D + 2)y = e2*(1 + e2x)¯* -

Given:

$\left({D}^{2}-3D+2\right)y={e}^{2x}{\left(1+{e}^{2x}\right)}^{-1}$

Formula used:
$y={y}_{h}+{y}_{p}$

$\left({D}^{2}-3D+2\right)y={e}^{2x}{\left(1+{e}^{2x}\right)}^{-1}$

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=yh+yp

Where,

• yh is the solution to the homogeneous ODE $\left({a}_{n}{D}^{n}+......+{a}_{1}D+{a}_{0}\right)y=0$
• yp the particular solution is any function that satisfies the non-homogeneous equation

On finding yh by solving (D2-3D+2)y=0

y=c1e2x+c2ex

On finding yp that satisfies (D2-3D+2)y=e2x(1+e2x)-1

Therefore the general solution y=yh+yp is

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