General Solution of Higher Order PDEs with Constant Coefficients - CSIR-NET Mathematical Sciences | Mathematics for IIT JAM, GATE, CSIR NET, UGC NET PDF Download

 Page 1


PARTIAL DIFFERENTIAL EQUATIONS OF HIGHER ORDER WITH CONSTANT
COEFFICIENTS.
 
Homogeneous Linear Equations with constant Coefficients .
A homogeneous linear partial differential equation of the n order is of the form
 
homogeneous because all its terms contain derivatives of the same order.
Equation (1) can be expressed as
 
As in the case of ordinary linear equations with constant coefficients the complete solution of (1) consists of two parts, namely, the
complementary function and the particular integral.
 
The complementary function is the complete solution of f (D,D ) z = 0-------(3), which must contain n arbitrary functions as the degree
of the polynomial f(D,D ). The particular integral is the particular solution of equation (2).
 
Finding the complementary function
 
Let us now consider the equation f(D,D ) z = F (x,y)
 
The auxiliary equation of (3) is obtained by replacing D by m and D by 1.
 
Solving equation (4) for „m?, we get „n? roots. Depending upon the nature of the roots, the Complementary function is written as given
below:
 
 
 
Finding the particular Integral
th
'
'
'
'
Page 2


PARTIAL DIFFERENTIAL EQUATIONS OF HIGHER ORDER WITH CONSTANT
COEFFICIENTS.
 
Homogeneous Linear Equations with constant Coefficients .
A homogeneous linear partial differential equation of the n order is of the form
 
homogeneous because all its terms contain derivatives of the same order.
Equation (1) can be expressed as
 
As in the case of ordinary linear equations with constant coefficients the complete solution of (1) consists of two parts, namely, the
complementary function and the particular integral.
 
The complementary function is the complete solution of f (D,D ) z = 0-------(3), which must contain n arbitrary functions as the degree
of the polynomial f(D,D ). The particular integral is the particular solution of equation (2).
 
Finding the complementary function
 
Let us now consider the equation f(D,D ) z = F (x,y)
 
The auxiliary equation of (3) is obtained by replacing D by m and D by 1.
 
Solving equation (4) for „m?, we get „n? roots. Depending upon the nature of the roots, the Complementary function is written as given
below:
 
 
 
Finding the particular Integral
th
'
'
'
'
 
Expand [f (D,D )] in ascending powers of D or D and operate on x y term by term.
 
Case (iv) : When F(x,y) is any function of x and y.
 
into partial fractions considering f (D,D ) as a function of D alone.
Then operate each partial fraction on F(x,y)  in such a way that 
 
where c is replaced by y+mx after integration
 
Example 26
 
Solve(D –3D D + 4D ) z = e
 
The auxillary equation is m=m –3m + 4 = 0
The roots are m = -1,2,2
 
Therefore the C.F is f (y-x) + f (y+ 2x) + xf (y+2x).
' -1 ' m n
'
3 2 ' '3 x+2y
3 2
1 2 3
Page 3


PARTIAL DIFFERENTIAL EQUATIONS OF HIGHER ORDER WITH CONSTANT
COEFFICIENTS.
 
Homogeneous Linear Equations with constant Coefficients .
A homogeneous linear partial differential equation of the n order is of the form
 
homogeneous because all its terms contain derivatives of the same order.
Equation (1) can be expressed as
 
As in the case of ordinary linear equations with constant coefficients the complete solution of (1) consists of two parts, namely, the
complementary function and the particular integral.
 
The complementary function is the complete solution of f (D,D ) z = 0-------(3), which must contain n arbitrary functions as the degree
of the polynomial f(D,D ). The particular integral is the particular solution of equation (2).
 
Finding the complementary function
 
Let us now consider the equation f(D,D ) z = F (x,y)
 
The auxiliary equation of (3) is obtained by replacing D by m and D by 1.
 
Solving equation (4) for „m?, we get „n? roots. Depending upon the nature of the roots, the Complementary function is written as given
below:
 
 
 
Finding the particular Integral
th
'
'
'
'
 
Expand [f (D,D )] in ascending powers of D or D and operate on x y term by term.
 
Case (iv) : When F(x,y) is any function of x and y.
 
into partial fractions considering f (D,D ) as a function of D alone.
Then operate each partial fraction on F(x,y)  in such a way that 
 
where c is replaced by y+mx after integration
 
Example 26
 
Solve(D –3D D + 4D ) z = e
 
The auxillary equation is m=m –3m + 4 = 0
The roots are m = -1,2,2
 
Therefore the C.F is f (y-x) + f (y+ 2x) + xf (y+2x).
' -1 ' m n
'
3 2 ' '3 x+2y
3 2
1 2 3
 
Hence, the solution is z  = C.F. + P.I
 
 
Example 27
Solve (D –4DD +4 D ) z  = cos (x –2y)
The auxiliary equation is m –4m + 4 = 0
Solving, we get m = 2,2.
Therefore the        C.F is f1(y+2x) + xf2(y+2x).
 
 
Example 28
 
Solve (D –2DD ) z = x y + e
 
The auxiliary equation is m –2m = 0.
Solving, we get     m = 0,2.
Hence the    C.F    is       f1 (y) + f2 (y+2x).
2 ' ' 2
2
2 ' 3 5x
2
Page 4


PARTIAL DIFFERENTIAL EQUATIONS OF HIGHER ORDER WITH CONSTANT
COEFFICIENTS.
 
Homogeneous Linear Equations with constant Coefficients .
A homogeneous linear partial differential equation of the n order is of the form
 
homogeneous because all its terms contain derivatives of the same order.
Equation (1) can be expressed as
 
As in the case of ordinary linear equations with constant coefficients the complete solution of (1) consists of two parts, namely, the
complementary function and the particular integral.
 
The complementary function is the complete solution of f (D,D ) z = 0-------(3), which must contain n arbitrary functions as the degree
of the polynomial f(D,D ). The particular integral is the particular solution of equation (2).
 
Finding the complementary function
 
Let us now consider the equation f(D,D ) z = F (x,y)
 
The auxiliary equation of (3) is obtained by replacing D by m and D by 1.
 
Solving equation (4) for „m?, we get „n? roots. Depending upon the nature of the roots, the Complementary function is written as given
below:
 
 
 
Finding the particular Integral
th
'
'
'
'
 
Expand [f (D,D )] in ascending powers of D or D and operate on x y term by term.
 
Case (iv) : When F(x,y) is any function of x and y.
 
into partial fractions considering f (D,D ) as a function of D alone.
Then operate each partial fraction on F(x,y)  in such a way that 
 
where c is replaced by y+mx after integration
 
Example 26
 
Solve(D –3D D + 4D ) z = e
 
The auxillary equation is m=m –3m + 4 = 0
The roots are m = -1,2,2
 
Therefore the C.F is f (y-x) + f (y+ 2x) + xf (y+2x).
' -1 ' m n
'
3 2 ' '3 x+2y
3 2
1 2 3
 
Hence, the solution is z  = C.F. + P.I
 
 
Example 27
Solve (D –4DD +4 D ) z  = cos (x –2y)
The auxiliary equation is m –4m + 4 = 0
Solving, we get m = 2,2.
Therefore the        C.F is f1(y+2x) + xf2(y+2x).
 
 
Example 28
 
Solve (D –2DD ) z = x y + e
 
The auxiliary equation is m –2m = 0.
Solving, we get     m = 0,2.
Hence the    C.F    is       f1 (y) + f2 (y+2x).
2 ' ' 2
2
2 ' 3 5x
2
 
 
 
Example 29
 
The auxiliary equation is m + m –6 = 0.
Therefore,   m = –3, 2.   
Hence the C.F is   f1(y-3x) + f2(y + 2x).
2
Page 5


PARTIAL DIFFERENTIAL EQUATIONS OF HIGHER ORDER WITH CONSTANT
COEFFICIENTS.
 
Homogeneous Linear Equations with constant Coefficients .
A homogeneous linear partial differential equation of the n order is of the form
 
homogeneous because all its terms contain derivatives of the same order.
Equation (1) can be expressed as
 
As in the case of ordinary linear equations with constant coefficients the complete solution of (1) consists of two parts, namely, the
complementary function and the particular integral.
 
The complementary function is the complete solution of f (D,D ) z = 0-------(3), which must contain n arbitrary functions as the degree
of the polynomial f(D,D ). The particular integral is the particular solution of equation (2).
 
Finding the complementary function
 
Let us now consider the equation f(D,D ) z = F (x,y)
 
The auxiliary equation of (3) is obtained by replacing D by m and D by 1.
 
Solving equation (4) for „m?, we get „n? roots. Depending upon the nature of the roots, the Complementary function is written as given
below:
 
 
 
Finding the particular Integral
th
'
'
'
'
 
Expand [f (D,D )] in ascending powers of D or D and operate on x y term by term.
 
Case (iv) : When F(x,y) is any function of x and y.
 
into partial fractions considering f (D,D ) as a function of D alone.
Then operate each partial fraction on F(x,y)  in such a way that 
 
where c is replaced by y+mx after integration
 
Example 26
 
Solve(D –3D D + 4D ) z = e
 
The auxillary equation is m=m –3m + 4 = 0
The roots are m = -1,2,2
 
Therefore the C.F is f (y-x) + f (y+ 2x) + xf (y+2x).
' -1 ' m n
'
3 2 ' '3 x+2y
3 2
1 2 3
 
Hence, the solution is z  = C.F. + P.I
 
 
Example 27
Solve (D –4DD +4 D ) z  = cos (x –2y)
The auxiliary equation is m –4m + 4 = 0
Solving, we get m = 2,2.
Therefore the        C.F is f1(y+2x) + xf2(y+2x).
 
 
Example 28
 
Solve (D –2DD ) z = x y + e
 
The auxiliary equation is m –2m = 0.
Solving, we get     m = 0,2.
Hence the    C.F    is       f1 (y) + f2 (y+2x).
2 ' ' 2
2
2 ' 3 5x
2
 
 
 
Example 29
 
The auxiliary equation is m + m –6 = 0.
Therefore,   m = –3, 2.   
Hence the C.F is   f1(y-3x) + f2(y + 2x).
2
  
 
=   ò(c + 3x) d(–cosx) –2 òcosx  dx 
 
= (c + 3x) (–cosx) –(3) ( - sinx) –2 sinx 
 
=   –y cosx + sinx 
 
Hence the complete solution is
 
z = f (y –3x) + f (y + 2x) –y cosx + sinx
 
Example 30
Solve r –4s + 4t  2x +y
 
i.e, (D –4DD + 4D  ) z = e
The auxiliary equation is m –4m + 4 = 0.
Therefore, m = 2,2
Hence the C.F is f (y + 2x) + x f (y + 2x).
 
Since D –4DD +4D = 0 for D = 2  and D = 1, we have to apply the general rule.
1 2
= e
2 ' ' 2 2x + y
2
1 2
2 ' '2 '