fract{d~y,d~x} _ = _ f ( ~x ) g ( ~y )
int{array{/},,,}fract{1 ,g ( ~y )} d~y _ _ = _ _ int{array{/},,,} f ( ~x ) d~x _ + _ ~c
fract{d~y,d~x} _ = _ f rndb{fract{~y,~x}}
Put _ ~v ( ~x ) _ = _ ~y ./ ~x , _ i.e. _ ~y _ = _ ~v ~x
fract{d~y,d~x} _ _ = _ _ ~v _ + _ ~x fract{d~v,d~x} _ _ = _ _ f ( ~v )
fract{d~v,d~x} _ _ = _ _ fract{f ( ~v ) - ~v,~x}
int{array{/},,,}fract{1 ,f ( ~v ) - ~v} d~v _ _ = _ _ int{array{/},,,}fract{1,~x} d~x _ + _ ~c
log ~x _ + _ c _ _ = _ _ int{array{/},,,}fract{1 ,f ( ~v ) - ~v} d~v
#{Example}
fract{d~y,d~x} _ = _ fract{~y - ~x + 1,~y + ~x + 5}
Put _ ~X = ~x + ~a, _ ~Y = ~y + ~b , _ then
fract{d~y,d~x} _ = _ fract{d~Y,d~X} _ = _ fract{~Y - ~b - ~X + ~a + 1,~Y - ~b + ~X - ~a + 5}
Now find ~a, ~b , such that _ - ~b + ~a + 1 _ = _ 0 , _ and _ - ~b + - ~a + 5 _ = _ 0 ,
i.e. _ ~a = 2 , _ ~b = 3 . _ So
fract{d~Y,d~X} _ = _ fract{~Y - ~X,~Y + ~X} _ = _ fract{( ~Y ./ ~X ) - 1,( ~Y ./ ~X ) + 1} , _ _ where _ ~X = ~x + 2, _ ~Y = ~y + 3
Put _ ~v _ = _ ~Y ./ ~X , _ we get
~X fract{d~v,d~X} _ = _ fract{ ~v - 1, ~v + 1} - ~v _ = _ fract{ - ( ~v^2 + 1 ), ~v + 1}
log ~X _ + _ ~c _ _ = _ _ - int{array{/},,,}fract{~v + 1,~v^2 + 1} d~v _ _ = _ _ - int{array{/},,,}fract{~v,~v^2 + 1} d~v - int{array{/},,,}fract{1,~v^2 + 1} d~v
_
_ _ _ _ _ log ~X _ + _ ~c _ _ = _ _ - &hlf. log ( ~v^2 + 1 ) + tan ^{-1}(~v)
_ _ _ _ _ log ~X ^2 + 2~c + log ( (~Y/~X)^2 + 1 ) - 2 tan ^{-1}(~Y/~X) _ _ = _ _ 0
_ _ _ _ _ 2~c + log ( ~Y ^2 + ~X ^2 ) - 2 tan ^{-1}(~Y/~X) _ _ = _ _ 0
Giving the solution:
log ( ( ~y + 3 )^2 + ( ~x + 2 )^2 ) _ - _ 2 tan ^{-1} rndb{fract{~y + 3,~x + 2}} _ + _ ~k _ _ = _ _ 0
_ ~f(~x) fract{d~y,d~x} + ~g(~x) ~y + ~h(~x) _ = _ 0
Normalizing:
fract{d~y,d~x} + ~P(~x) ~y = ~Q(~x)
find ~R such that _ ~R ~P = ^{d~R}/_{d~x} , _ and multiply:
~R fract{d~y,d~x} + ~R ~P ~y = ~R ~Q
~R fract{d~y,d~x} + ~y fract{d~R,d~x} = ~R ~Q
fract{d ~y ~R,d~x} _ = _ ~R ~Q
_ _ _ _ _ ~y ~R _ = _ ~{∫} ~R ~Q d~x + ~c
To find ~R , _ ~R ~P = ^{d~R}/_{d~x} _ <= _ ~{∫} ^1/_~R d~R = ~{∫} ~P d~x _ <= _ ln ~R = ~{∫} ~P d~x _ <= _ ~R = exp \{~{∫} ~P d~x\}
#{Example}
^{d~y}/_{d~x} + λ ~y = &alpha. - &beta. ~x
Try _ ~R = ~e^{λ~x}
~e^{λ~x} _ + ~e^{λ~x} λ ~y _ = _ ~e^{λ~x} ( &alpha. - &beta. ~x )
~y ~e^{λ~x} _ = _ ~{∫} ~e^{λ~x} ( &alpha. - &beta. ~x ) d~x _ = _ &alpha. ~{∫} ~e^{λ~x} d~x - &beta. ~{∫} ~x ~e^{λ~x} d~x
^d/_{d~x} ( ( λ ~x ~e^{λ ~x} - ~e^{λ ~x} ) _ = _ λ^2 ~x ~e^{λ ~x} + λ ~e^{λ ~x} - λ ~e^{λ ~x} _ = _ λ^2 ~x ~e^{λ ~x}