Question

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

Answer 1

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.