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Name: How to solve au¨+b(u˙)2+u˙+u˙ceu+eu−e2u+1=0
Uploaded: 2022-02-23T17:22:57+00:00
Description: How to solve au¨+b(u˙)2+u˙+u˙ceu+eu−e2u+1=0

eozoischgc

Answered question

2022-01-19

How to solve

a \ddot{u} + b {(\dot{u})}^{2} + \dot{u} + \dot{u} c e^{u} + e^{u} - e^{2 u} + 1 = 0

Answer & Explanation

Lynne Trussell

Beginner2022-01-19Added 32 answers

a \ddot{u} + b {(\dot{u})}^{2} + \dot{u} + \dot{u} c e^{u} + e^{u} - e^{2 u} + 1 = 0

a \frac{d^{2} u}{{d t}^{2}} + b {(\frac{d u}{d t})}^{2} + (c e^{u} + 1) \frac{d u}{d t} + e^{u} - e^{2 u} + 1 = 0

Case 1:

a \neq 0

Let

v = \frac{d u}{d t}

,
Then

\frac{d^{2} u}{{d t}^{2}} = \frac{d v}{d t} = \frac{d v}{d u} \frac{d u}{d t} = v \frac{d v}{d u}

∴ a v \frac{d v}{d u} + b v^{2} + (c e^{u} + 1) v + e^{u} - e^{2 u} + 1 = 0

a v \frac{d v}{d u} = - b v^{2} - (c e^{u} + 1) v + e^{2 u} - e^{u} - 1

v \frac{d v}{d u} = - \frac{b v^{2}}{a} - \frac{(c e^{u} + 1) v}{a} + \frac{e^{2 u} - e^{u} - 1}{a}

This belongs to an Abel equation of the second kind.
In fact, all Abel equation of the second kind can be transformed into Abel equation of the first kind.
Let

v = \frac{1}{w}

,
Then

\frac{d v}{d u} = - \frac{1}{w^{2}} \frac{d w}{d u}

∴ - \frac{1}{w^{3}} \frac{d w}{d u} = - \frac{b}{a w^{2}} - \frac{c e^{u} + 1}{a w} + \frac{e^{2 u} - e^{u} - 1}{a}

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