# Second Order Differential Equations

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## General Second Order Linear Equation

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.

### Complementary Function

The solution, ~y_1, of the corresponding homogenous equation , _ ~Q ( ~y ) _ = _ 0 , _ is called the #~{complementary function}.

### Particular Integral

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} - ***