In my differential equations book, I have found the following:Let P 0 ( d y d x ) n + P 1 ( d y d x ) n − 1 + P 2 ( d y d x ) n − 2 + . . . . . . + P n − 1 ( d y d x ) + P n = 0 be the differential equation of first degree 1 and order n (where P i ∀ i ∈ 0 , 1 , 2 , . . . n are functions of x and y).Assuming that it is solvable for p, it can be represented as: [ p − f 1 ( x , y ) ] [ p − f 2 ( x , y ) ] [ p − f 3 ( x , y ) ] . . . . . . . . [ p − f n ( x , y ) ] = 0equating each factor to Zero, we get n differential equations of first order and first degree. [ p − f 1 ( x , y ) ] = 0 , [ p − f 2 ( x , y ) ] = 0 , [ p − f 3 ( x , y ) ] = 0 , . . . . . . . . [ p − f n ( x , y ) ] = 0Let the solution to these n factors be: F 1 ( x , y , c 1 ) = 0 , F 2 ( x , y , c 2 ) = 0 , F 3 ( x , y , c 3 ) = 0 , . . . . . . . . F n ( x , y , c n ) = 0Where c 1 , c 2 , c 3 . . . . . c n are arbitrary constants of integration. Since all the c’s can have any one of an infinite number of values, the above solutions will remain general if we replace c 1 , c 2 , c 3 . . . . . c n by a single arbitrary constant c. Then the n solutions (4) can be re-written as F 1 ( x , y , c ) = 0 , F 2 ( x , y , c ) = 0 , F 3 ( x , y , c ) = 0 , . . . . . . . . F n ( x , y , c ) = 0They can be combined to form the general solution as follows: F 1 ( x , y , c ) F 2 ( x , y , c ) F 3 ( x , y , c ) . . . . . . . . F n ( x , y , c ) = 0 ( 1 )Now, my question is, whether equation (1) is the most general form of solution to the differential equation.I think the following is the most general form of solution to the differential equation : F 1 ( x , y , c 1 ) F 2 ( x , y , c 2 ) F 3 ( x , y , c 3 ) . . . . . . . . F n ( x , y , c n ) = 0 ( 2 )If (1) is the general solution, the constant of integration can be found out by only one IVP say, y ( 0 ) = 0. So, one IVP will give the particular solution. If (2) is the general solution, one IVP might not be able to give the particular solution to the problem.

Question

In my differential equations book, I have found the following:Let       P    0    (            d      y              d      x            )    n    +      P    1    (            d      y              d      x            )          n      −      1        +      P    2    (            d      y              d      x            )          n      −      2        +  .  .  .  .  .  .  +      P          n      −      1        (            d      y              d      x        )  +      P    n    =  0 be the differential equation of first degree 1 and order n (where       P    i     ∀ i   ∈      0    ,    1    ,    2    ,    .    .    .    n   are functions of x and y).Assuming that it is solvable for p, it can be represented as:  [  p  −      f    1    (  x  ,  y  )  ]  [  p  −      f    2    (  x  ,  y  )  ]  [  p  −      f    3    (  x  ,  y  )  ]  .  .  .  .  .  .  .  .  [  p  −      f    n    (  x  ,  y  )  ]  =  0equating each factor to Zero, we get n differential equations of first order and first degree.  [  p  −      f    1    (  x  ,  y  )  ]  =  0  ,     [  p  −      f    2    (  x  ,  y  )  ]  =  0  ,     [  p  −      f    3    (  x  ,  y  )  ]  =  0  ,     .  .  .  .  .  .  .  .  [  p  −      f    n    (  x  ,  y  )  ]  =  0Let the solution to these n factors be:      F    1    (  x  ,  y  ,      c    1    )  =  0  ,         F    2    (  x  ,  y  ,      c    2    )  =  0  ,         F    3    (  x  ,  y  ,      c    3    )  =  0  ,     .  .  .  .  .  .  .  .      F    n    (  x  ,  y  ,      c    n    )  =  0Where       c    1    ,      c    2    ,      c    3    .  .  .  .  .      c    n   are arbitrary constants of integration. Since all the c’s can have any one of an infinite number of values, the above solutions will remain general if we replace       c    1    ,      c    2    ,      c    3    .  .  .  .  .      c    n   by a single arbitrary constant c. Then the n solutions (4) can be re-written as      F    1    (  x  ,  y  ,  c  )  =  0  ,         F    2    (  x  ,  y  ,  c  )  =  0  ,         F    3    (  x  ,  y  ,  c  )  =  0  ,     .  .  .  .  .  .  .  .      F    n    (  x  ,  y  ,  c  )  =  0They can be combined to form the general solution as follows:      F    1    (  x  ,  y  ,  c  )         F    2    (  x  ,  y  ,  c  )         F    3    (  x  ,  y  ,  c  )     .  .  .  .  .  .  .  .      F    n    (  x  ,  y  ,  c  )  =  0                                      (  1  )Now, my question is, whether equation (1) is the most general form of solution to the differential equation.I think the following is the most general form of solution to the differential equation :      F    1    (  x  ,  y  ,      c    1    )         F    2    (  x  ,  y  ,      c    2    )         F    3    (  x  ,  y  ,      c    3    )     .  .  .  .  .  .  .  .      F    n    (  x  ,  y  ,      c    n    )  =  0                                      (  2  )If (1) is the general solution, the constant of integration can be found out by only one IVP say,   y  (  0  )  =  0. So, one IVP will give the particular solution. If (2) is the general solution, one IVP might not be able to give the particular solution to the problem.

stacan6t · Accepted Answer

Indeed equation 1 is the general solution! Notice that equation 1 is satisfied even if any one of       F    i    (  x  ,  y  ,      c    i    ) is zero.Suppose that you have   c  =      c    1  . Then equation 1 is satisfied because       F    1    =  0. If you have   c  =      c    2  , then again eq. 1 is satisfied because       F    2    =  0. And so on.Your equation 2 is a special solution, as in this case, all       F    1    ,      F    2    …      F    n   are individually zero.

In my differential equations book, I have found the following: Let P 0 </msub>

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