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# First Order Differential Equations

## Separable

fract{d~y,d~x} _ = _ f ( ~x ) g ( ~y )

int{array{/},,,}fract{1 ,g ( ~y )} d~y _ _ = _ _ int{array{/},,,} f ( ~x ) d~x _ + _ ~c

## Homogenous

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

## Reduction to Homogenous

#{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

## General First Order

_ ~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}