I've found this particular equation rather tough, can you give me some hints on how...

ntaraxq

ntaraxq

Answered

2022-07-03

I've found this particular equation rather tough, can you give me some hints on how to solve
y ˙ + t cos π y 2 + 1 t = y

Answer & Explanation

Zachery Conway

Zachery Conway

Expert

2022-07-04Added 7 answers

y ˙ + t cos π y 2 + 1 t = y
d y d t = ( 1 cos π y 2 ) t + y 1 y ˙ + t cos π y 2 + 1 t = y
Let u = y 1,
Then y = u + 1
d y d t = d u d t
d u d t = ( 1 cos π ( u + 1 ) 2 ) t + u
( ( 1 cos ( π u 2 + π 2 ) ) t + u ) d t d u = 1
( ( 1 + sin π u 2 ) t + u ) d t d u = 1
Let v = t + u 1 + sin π u 2 ,
Then t = v u 1 + sin π u 2
d t d u = d v d u + π u 2 cos π u 2 ( 1 + sin π u 2 ) 2 1 1 + sin π u 2
( 1 + sin π u 2 ) v ( d v d u + π u 2 cos π u 2 ( 1 + sin π u 2 ) 2 1 1 + sin π u 2 ) = 1
( 1 + sin π u 2 ) v d v d u + ( π u 2 cos π u 2 1 + sin π u 2 1 ) v = 1
( 1 + sin π u 2 ) v d v d u = ( 1 π u 2 cos π u 2 1 + sin π u 2 ) v + 1
v d v d u = ( 1 1 + sin π u 2 π u 2 cos π u 2 ( 1 + sin π u 2 ) 2 ) v + 1 1 + sin π u 2
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 = 1 w ,
Then d v d u = 1 w 2 d w d u
1 w 3 d w d u = ( 1 1 + sin π u 2 π u 2 cos π u 2 ( 1 + sin π u 2 ) 2 ) 1 w + 1 1 + sin π u 2
d w d u = w 3 1 + sin π u 2 + ( π u 2 cos π u 2 ( 1 + sin π u 2 ) 2 1 1 + sin π u 2 ) w 2

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