Question

find pi of the give pde D^3-4D^2D'+5DD'^2-2D'^3= e^(y+2x)+(x+y)^1/2

Answer 1

Solution:

The equation can be written as

(D2 – 5D + 6)y = e3x

(D – 3) (D – 2)y = e3x

C.F. = c1 e3x + c2e2x

And P.I. =1/ (D-3). 1/ (D-2) e3x

= 1/ (D-3) e2x ∫ e3x e-2xdx

=1/ (D-3) e2x ex

= e3x ∫ e3x e-3x dx = x.e3x

y = c1e3x +c2e2x +xe3x

P.I. can be found by resolving

1/f (D) = 1/ (D-3). 1/ (D-2)

Now using partial fractions,

1/f (D) = 1/ (D-3). 1/ (D-2)

= 1/ (D-3) – 1/ (D-2)

Hence, the required P.I. is [1(D-3) – 1/ (D-2)] e3x

= 1/ (D-3) e3x – 1/ (D-2) e3x

= e3x ∫ e3x e-3x dx – e2x ∫ e3x e-2x dx

= xe3x – e3x

Second term can be neglected as it is included in the first term of the C.F.