Page 127 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes In fact, Leibnitz s formula yields
p
( )
p
n
y (p) = y z ( ) py z (p 1)  y z (p 1,2,...., )
1
1
1
(n)
Substituting these in (5), we see that the coefficient of z is y , and that of z is zero. Thus (5)
1
becomes an equation of the (n 1) order with respect to z ,
n
x
x
y z ( ) q 1 ( )z (n 1) q 2 ( )z (n 1) ... q n 1 ( )z = 0 (vi)
x
1
In particular, when n = 2, the reduced equation (vi) can be solved. Hence, by virtue of this
method, we obtain the general solution
x i
2
t
x
y(x) = y 1 ( ) y 1 ( ) exp p 1 ( )d dt ...(viii)
y (x) being a particular solution of (5) with n = 2. This method is useful in the practical
1
treatment of the linear differential equations.
Self Assessment
1. Consider the second order differential equations
x
x
y p 1 ( )y p 2 ( )y 0
having two independent solutions y and y . Find a relation between p , p in terms of y ,
1 2 1 2 1
y and their derivatives.
2
2. Obtain the particular solution of the differential equation
y y e 2x
by the method of variation of constants.
7.3 Solution of the Linear Equation with Constant Coefficients

To solve the equation
n
d y d n 1 y
P P ... P y = 0, ...(i)
0 n 1 n 1 n
dx dx
where P , P ,...., P are constants.
0 1 n
mx
Substitute y = e on a trial basis,
Then e mx (P m n P m n 1 ... P ) = 0 ...(ii)
0 1 n
mx
Now, e is a solution of (i) if m is a root of the algebraic equation
P m n P m n 1 ... P = 0 ...(iii)
0 1 n
Auxiliary Equation

The equation (iii) is called the auxiliary equation. Therefore if m have a value say m that satisfies
1
(iii), y = e m 1 x is an integral of (i), and if the n roots of (iii) be m , m , m , ...m the complete solution
1 2 3 n
of (i) is
y = c e m 1 x c e m 2 x ... c e m n x .
1 2 n

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