The general second order linear differential equation has the form:
~a fract{d^2~y,d~x^2} + ~b fract{d~y,d~x} + ~c ~y _ = _ ~f ( ~x ) _ _ _ _ _ _ ~a, ~b, ~c _ constants.
For convenience we write:
~Q ( ~y ) _ = _ ~f ( ~x ) , _ where _ ~Q ( ~y ) _ = _ ~a fract{d^2~y,d~x^2} + ~b fract{d~y,d~x} + ~c ~y
Now if ~y_1 is the general solution of the homogenous equation _ ~Q ( ~y_1 ) _ = _ 0, and ~y_2 is one solution of the general equation _ ~Q ( ~y_2 ) _ = _ ~f ( ~x ) , _ then it is easy to see (by linearity of differential operator) that
~Q ( ~y_1 + ~y_2 ) _ = _ ~Q ( ~y_1 ) + ~Q ( ~y_2 ) _ = _ ~f ( ~x ) + 0 _ = _ ~f ( ~x )
So _ ~y_1 + ~y_2 _ is also a solution of the general equation, in fact this is the general solution.
The solution, ~y_1, of the corresponding homogenous equation , _ ~Q ( ~y ) _ = _ 0 , _ is called the #~{complementary function}.
Any solution, ~y_2, of the equation _ ~Q ( ~y_2 ) _ = _ ~f ( ~x ) _ is called a #~{particular integral} of the second order differential equation.
The technique is therefore to find the complementary function and a paricular integral, and take the sum.
#{Example 1}
fract{d^2~y,d~x^2} + 3 fract{d~y,d~x} - 54~y _ = _ 4 ~e ^{2~x}
The auxiliary equation _ ~m^2 + 3~m - 54 _ = _ 0 , _ has solutions _ ~p _ = _ 6 _ and _ ~q _ = _ -9 .
The complementary function is : _ ~y _ = _ ~A ~e ^{6~x} + ~B ~e ^{-9~x}
For a particular integral try _ ~y = ~c ~e ^{2~x} , _ ~y#~' = 2 ~c ~e ^{2~x}, _ ~y#~{''} = 4 ~c ~e ^{2~x}
~y#~{''} + 3 ~y#~' - 54 ~y _ = _ ( 4 + 6 - 54 ) ~c ~e ^{2~x} _ _ => _ _ ~c = - ^1/_{11}
So the general solution is : _ ~y _ = _ ~A ~e ^{6~x} + ~B ~e ^{-9~x} - ^1/_{11} ~e ^{2~x}
#{Example 2}
fract{d^2~y,d~x^2} - ~y _ = _ 2 + 5~x
The auxiliary equation _ ~m^2 - 1 _ = _ 0 , _ has solutions _ ~p _ = _ 1 , _ ~q _ = _ -1 .
The complementary function is : _ ~y _ = _ ~A ~e ^{~x} + ~B ~e ^{-~x}
For a particular integral try _ ~y = &alpha. + &beta.~x , _ ~y#~' = &beta., _ ~y#~{''} = 0
~y#~{''} - ~y _ = _ -&alpha. - &beta.~x _ _ => _ _ &alpha. _ = _ -2 , _ &beta. _ = _ -5
So the general solution is : _ ~y _ = _ ~A ~e ^{~x} + ~B ~e ^{-~x} - 2 - 5~x
#{Example 3}
fract{d^2~y,d~x^2} + 3 fract{d~y,d~x} - 54~y _ = _ ~e ^{6~x}
The complementary function (as example 1) is : _ ~y _ = _ ~A ~e ^{6~x} + ~B ~e ^{-9~x}
For a particular integral try _ ~y = ~c ~e ^{6~x} , _ ~y#~' = 6 ~c ~e ^{6~x}, _ ~y#~{''} = 36 ~c ~e ^{6~x}
~y#~{''} + 3 ~y#~' - 54 ~y _ = _ (36 + 18 - 54) ~c ~e ^{6~x} _ = _ 0 ~c ~e ^{6~x} ... !
This will not give a solution so try _ ~y = ~x ~c ~e ^{6~x} , _ ~y#~' = ( 6~x + 1 ) ~c ~e ^{6~x} , _ ~y#~{''} = ( 36~x + 12 ) ~c ~e ^{6~x}
~y#~{''} + 3 ~y#~' - 54 ~y _ = _ ( 36~x + 12 + 18~x + 3 - 54~x) ~c ~e ^{6~x} _ = _ 15~c ~e ^{6~x} _ = _ ~e ^{6~x} _ _ => _ _ ~c _ = _ ^1/_{15}
So the general solution is : _ ~y _ = _ ~A ~e ^{6~x} + ~B ~e ^{-9~x} + ^1/_{15} ~e ^{6~x}
#{Example 4}
fract{d^2~y,d~x^2} - 3 fract{d~y,d~x} + 2 ~y _ = _ 2 sin ~x
The auxiliary equation _ ~m^2 - 3 ~m + 2 _ = _ 0 , _ has solutions _ ~p _ = _ 2 _ and _ ~q _ = _ 1 .
The complementary function is : _ ~y _ = _ ~A ~e ^{2~x} + ~B ~e ^{~x}
For a particular integral try _ ~y = ~C sin ~x + ~D cos ~x , _ ~y#~' = ~C cos ~x - ~D sin ~x, _ ~y#~{''} = -~C sin ~x - ~D cos ~x
~y#~{''} - 3 ~y#~' + 2 ~y _ = _ ( ~C + 3~D ) sin ~x + ( ~D - 3~C ) cos ~x _ = _ 2 sin ~x _ _ => _ _ ~C = 0.2, _ ~D = 0.6
So the general solution is : _ ~y _ = _ ~A ~e ^{2~x} + ~B ~e ^{~x} + 0.2 sin ~x + 0.6 cos ~x
#{Example 5}
fract{d^2~y,d~x^2} + ~y _ = _ 4 cos ~x
The complementary function is : _ ~y _ = _ ~A sin ~x + ~B cos ~x . _ [See homogenous equations example 3 .]
So _ ~y = ~C sin ~x + ~D cos ~x _ is the same as the complementary function, and cannot therefore be used.
Instead try _ ~y = ~x ( ~C sin ~x + ~D cos ~x )
~y#~{''} + ~y _ = _ _ _ => _ _ *** = - ***
So the general solution is : _ ~y _ = _ ~A ~e ^{*~x} + ~B ~e ^{*~x} - ***
#{Example 6}
fract{d^2~y,d~x^2} - ~y _ = _ 2 + 5~x + sin 3~x
The auxiliary equation (as example 2), has solutions _ ~p _ = _ 1 _ and _ ~q _ = _ -1 .
The complementary function is : _ ~y _ = _ ~A ~e ^{*~x} + ~B ~e ^{*~x}
For a particular integral try _ ~y = *** , _ ~y#~' = ***, _ ~y#~{''} = ***
~y#~{''} + * ~y#~' + * ~y _ = _ _ _ => _ _ *** = - ***
So the general solution is : _ ~y _ = _ ~A ~e ^{*~x} + ~B ~e ^{*~x} - ***