Identify the following differential equations and hence solve it y=xy'+1-ln y'?

1 Answer
Jul 3, 2018

Answer:

General Solution:

  • # bb( y = c x + 1 - ln c )#

and singular solution:

  • #y = 2 + ln x#

Explanation:

This is in the form of Clairaut's equation:

#y = y' x + underbrace(F(y'))_(= 1 - ln y') qquad square#

Differentiate wrt #x#:

#cancel(y') = cancel(y') + y'' x + (dF)/(d y') y''#

And factor:

#:. y'' ( x + (dF)/(d y') ) = 0 implies {(y'' = 0 qquad qquad bbX),((dF)/(d y') = - x qquad bbY):}#

  • #bbX#

#y'' = 0 implies {(y' = c_1 ),(y = c_1 x + c_2 ):} qquad star#

Subbing both #star# equations into #square#:

# c_1 x + c_2 = c_1 x + F(c_1) implies c_2 = F(c_1)#

This gives the general solution:

# bb( y = c x + 1 - ln c )#

  • #bbY#

#d/(dy') (1 - ln y') = - x qquad implies -1/(y') = -x#

#implies y' = 1/x#

Putting that into #square#:

#y = 1/x x + 1 - ln (1/x)#

#y = 2 + ln x#

This is a singular solution.