Question

The Formula for finding Particular integral
of the function 1/(D-a)^r × e^ax is​

Answer 1

Answer:

Step-by-step explanation: Particular Integral

[1/f(D)]eax = [1/f(a)]eax

If f(a) = 0 then [1/f(D)]eax = x[1/f'(a)]eax

If f'(a) = 0 then [1/f(D)]eax = x2[1/f''(a)]eax

[1/f(D)]xn = [f(D)]-1xn expand [f(D)]-1 and then operate

[1/f(D2)]sin ax = [1/f(-a2)]sin ax

and [1/f(D2)]cos ax = [1/f(-a2)]cos ax

If f(-a2) = 0 then [1/f(D2)]sin ax = x[1/f'(-a2)]sin ax

Similar argument for cos.

[1/f(D)]eax φ(x) = eax [1/f(D+a)]φ(x)

[1/(D+a)]φ(x) = e-ax∫eaxφ(x) dx

Particular integral of product of polynomial and trigonometric functions

[1/f(D)]xn sin ax

[1/f(D)]xn(cos ax + i sin ax) = [1/f(D)]xn eiax = eiax[1/f(D+ia)]xn

[1/f(D)]xn sin ax = Imaginary part of eiax [1/f(D+ia)]xn

[1/f(D)]xn cos ax = Real part of eiax [1/f(D+ia)]xn

General Method of finding particular integral

General method of finding the particular integral of any function φ(x)

P.I. = [1/(D - a)]φ(x) = y

or

(D -  a)/(D - a) φ(x) = (D - a) y

φ(x) = (D - a) y or φ(x) = Dy - ay

dy/dx - ay =  φ(x) which is a linear differential equation.

Its solution is ye-∫adx = ∫e-∫adxφ(x)dx  

or ye-ax = ∫e-axφ(x)dx

y = eax∫e-axφ(x)dx

[1/(D - a)]φ(x) = eax∫e-axφ(x)dx