Once we are quiet familiar with several methods for solving an ordinary differential equation of first order and first degree, we can comfortably solve differential equations of second order and higher orders in first degree.

The solutions of a D.E of first order and first degree is arrived with direct involvement of integration, whereas integration is not directly involved in practical problems while arriving at the solution of the higher order equation. We also notice that the solution of a first order and first degree equation is mostly in the implicit form f ( x, y ) = c. In the case of higher order equations the solution is always in the explicit form y = f ( x ).

In this section let us discuss with higher order differential equations and the method of solving the higher order differential equations.

We know that the differential equations are those which consist differential operator, $\frac{\mathrm{d} }{\mathrm{d} x}$, $\frac{\mathrm{d^{2}x} }{\mathrm{d} x^{2}}$, etc.

Let us discuss with some of the differential equations and their orders.
1. $\frac{\mathrm{d^{2}y} }{\mathrm{d} x^{2}}$ + a1 $\frac{\mathrm{d}y }{\mathrm{d} x}$ + a2 y = 0

This is the homogeneous differential equation of order two, degree one, called linear second order differential equation.

2. $\frac{\mathrm{d^{3}y} }{\mathrm{d} x^{3}}$ + a1 $\frac{\mathrm{d^{2}y} }{\mathrm{d} x^{2}}$ + a2 $\frac{\mathrm{d}y }{\mathrm{d} x}$ + a3 y = g ( x )

This is the non-homogeneous differential equation of order three and degree one, called third order differential equation.

3. ( $\frac{\mathrm{d^{3}y} }{\mathrm{d} x^{3}}$ )2 + a1 ( $\frac{\mathrm{d^{2}y} }{\mathrm{d} x^{2}}$ )3 + a2 $\frac{\mathrm{d} y}{\mathrm{d} x}$ + a3 y = 0

The above equation is a homogeneous differential equation or order three and degree two.

4. $\frac{\mathrm{d^{5}y} }{\mathrm{d} x^{5}}$ + a1 ( $\frac{\mathrm{d^{4}y}}{\mathrm{d} x^{4}}$ )2 + a2 ( $\frac{\mathrm{d^{2}y} }{\mathrm{d} x^{2}}$ )3 + a3 $\frac{\mathrm{d} y }{\mathrm{d} x}$ + a3 y = g ( x )

The above equation is a non-homogeneous differential equation of 5th order and degree 1, since the degree of highest order is one.

This is also called as linear non-homogenous equation or order five.
The Differential Equation is said to be linear, if the dependent variable and its derivatives are of first degree. The differential equations will have two types of functions depending on whether it is homogeneous or non-homogeneous.For a homogeneous equations, the solution will be, y = Complimentary Function ( C. F. or yc ), which is found by writing the auxiliary equation of the given differential equation.

For a non-homogeneous differential equation we We have two different types of solutions called Complimentary Function ( yc ) and Particular Integral ( yp ).

So that the required solution will be y = yc + yp

Let us see some of the complimentary functions according to the roots of auxiliary equations.


Roots of Auxiliary Equations
Complimentary Functions ( yc )
1
3, 5
c1 e3x + c2 e5x
2
2, -2, 3, -3
c1 e2x + c2 e-2x + c3 e3x + c4 e-3x
3
5, 5 ( c1 + c2 x ) e3x
4
3, 3, 3, 0 ( c1 + c2 x c3 x2 ) + c4
5
$\pm$ 5i c1 cos 5x + c2 sin 5x
6
2 $\pm$ 3i e2x ( cos 3x + sin 3x )
7
2, 2, 1 $\pm$ 5i ( c1 + c2 x )e2x + ex ( cos 5x + sin 5x )
8
2 $\pm$ 3i , 2 $\pm$ 3i ex [ ( c1 + c2 x ) cos 2x + ( c3 + c4 x ) sin 2x ]

Working Procedure for problems to solve homogeneous differential equations with constant coefficients.

1. The given Differential Equation is put in the form, D y = 0

2. For the Auxiliary Equation, f ( r ) = 0 and solve the same ( here we have to use various techniques like, rational roots theorem, factor theorem and synthetic division).

3. Based on the nature of the roots of the Auxiliary Equation we write the Complementary Function ( yc ) which itself is the general solution of the D. E. taking into account the dependent and independent variables involved in the D.E.

Solved Example

Question: Solve the D. E $\frac{\mathrm{d^{3}y} }{\mathrm{d} x^{3}}$ + 6 $\frac{\mathrm{d^{2}y} }{\mathrm{d} x^{2}}$ + 11 $\frac{\mathrm{d}y }{\mathrm{d} x}$ + 6 y = 0
Solution:
 
We have $\frac{\mathrm{d^{3}y} }{\mathrm{d} x^{3}}$ + 6 $\frac{\mathrm{d^{2}y} }{\mathrm{d} x^{2}}$ + 11 $\frac{\mathrm{d}y }{\mathrm{d} x}$ + 6 y = 0

            Replacing $\frac{d}{dx}$ as D, the above D. E. can be rewritten as
                                       ( D3 + 6 D2 + 11 D + 6 ) y = 0

            The Auxiliary Equation is r3 + 6 r2 + 11 r + 6 = 0

   The possible rational factors are, $\pm$ 1, $\pm$ 2, $\pm$ 3, $\pm$ 6
                                                               Let f ( r ) = r3 + 6 r2 + 11 r + 6
                                                                  f ( - 1 ) = -1 + 6 - 11 + 6 = 0
                                                 =>                     x = -1 is one of the root of the auxiliary equation.
               Applying Synthetic division for the other roots,

            -1  |     1          6          11        6
                 |     0          -1         -5        -6   
                       1          5           6      |   0  

                              The quotient will be r2 + 5 x + 6 = 0
                               =>                  ( r + 2 ) ( r + 3 ) = 0
                               =>                                       r = -2, r = -3
            The real roots of the auxiliary equation are, r = -1, r = -2 and r = -3

                                     The General solution is y = c1 e-x + c2 e-2x + c3 e-3x

 


Solved Examples

Question 1: Solve the D. E. $\frac{\mathrm{d^{3}y} }{\mathrm{d} x^{3}}$ + y = 0
Solution:
 
We have   $\frac{\mathrm{d^{3}y} }{\mathrm{d} x^{3}}$ + y = 0
                                       The above D. E can be written as ( D3 + 1 ) y = 0
                  
                             The Auxiliary Equation is r3 + 1 = 0
                              =>             ( r + 1 ) ( r2 - r + 1 ) = 0
                              =>                                        r = -1
                                                                    and r = $\frac{1\pm\sqrt{1-4}}{2}$

                                                                            = $\frac{1\pm\sqrt{-3}}{2}$

                                                                            = $\frac{1\pm\sqrt{3}i^{2}}{2}$

                                                                            = $\frac{1}{2}$ $\pm$ i $\frac{\sqrt{3}}{2}$

                           The General Solution will be, y = c1 e-x  + ex/2 [ c2 cos ( $\frac{\sqrt{3}}{2}$ x ) + c3 sin ( $\frac{\sqrt{3}}{2}$ x ) ]
 

Question 2: $\frac{\mathrm{d^{4}y}}{\mathrm{d} x^{4}}$ -  5 $\frac{\mathrm{d^{2}y} }{\mathrm{d} x^{2}}$ + 4 y = 0
Solution:
 
We have, $\frac{\mathrm{d^{4}y}}{\mathrm{d} x^{4}}$ - 5 $\frac{\mathrm{d^{2}y} }{\mathrm{d} x^{2}}$ + 4 y = 0

           The above D. E can be written as ( D4 - 5 D2 + 4 ) y = 0

                           The Auxiliary Equation is, ( r4 - 5 r2 + 4 ) = 0
                              =>                             r4 -  4 r2 - r2 + 4 = 0
                              =>                     r2 ( r2 - 4 ) - 1 ( r2 - 4 ) = 0
                              =>                             ( r2 - 4 ) ( r2 - 1 ) = 0
                              =>        ( r + 2 ) ( r - 2 ) ( r + 1 ) ( r - 1 ) = 0
                              =>                                                   r = 2, - 2, 1, -1

                                        The general solution will be, y = c1 ex + c2 e-x + c3 e2x + c4 e-2x