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Chapter: Mathematics (maths) - Partial Differential Equations

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Partial Differential Equations of Higher Order With Constant Coefficients

Homogeneous Linear Equations with constant Coefficients. A homogeneous linear partial differential equation of the nth order is of the form

PARTIAL DIFFERENTIAL EQUATIONS OF HIGHER ORDER WITH CONSTANT COEFFICIENTS.

 

Homogeneous Linear Equations with constant Coefficients.

A homogeneous linear partial differential equation of the nth 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



Expand [f (D,D')]-1 in ascending powers of D or D' and operate on xm yn 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(D3 –3D2D' + 4D'3) z = ex+2y

 

The auxillary equation is m=m3 –3m2 + 4 = 0

The roots are m = -1,2,2

 

Therefore the C.F is f1(y-x) + f2 (y+ 2x) + xf3 (y+2x).



Hence, the solution is z  = C.F. + P.I


 

Example 27

Solve (D2 –4DD' +4 D' 2) z  = cos (x –2y)

The auxiliary equation is m2 –4m + 4 = 0

Solving, we get m = 2,2.

Therefore the        C.F is f1(y+2x) + xf2(y+2x).


 

Example 28

 

Solve (D2 –2DD') z = x3y + e5x

 

The auxiliary equation is m2 –2m = 0.

Solving, we get     m = 0,2.

Hence the    C.F    is       f1 (y) + f2 (y+2x).




 

Example 29


The auxiliary equation is m2 + m –6 = 0.

Therefore,   m = –3, 2.   

Hence the C.F is   f1(y-3x) + f2(y + 2x).

 

 

=   ò(c + 3x) d(–cosx) –2òcosx  dx 

 

= (c + 3x) (–cosx) –(3) ( - sinx) –2 sinx 

 

=   –y cosx + sinx 

 

Hence the complete solution is

 

z = f1(y –3x) + f2(y + 2x) –y cosx + sinx

 

Example 30

Solve r –4s + 4t = e 2x +y


i.e, (D2 –4DD' + 4D' 2  ) z = e2x + y

The auxiliary equation is m2 –4m + 4 = 0.

Therefore, m = 2,2

Hence the C.F is f1(y + 2x) + x f2(y + 2x).


Since D2 –4DD'+4D'2 = 0 for D = 2  and D' = 1, we have to apply the general rule.


 

 


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