Question

1 *
if Q(x) = e1007 V (x) then P.I​

Answer 1

Answer:

Step-by-step explanation:

If both m1 and m2 are constants, the expressions (D – m1) (D – m2) y and (D– m2) (D – m1) y are equivalent i.e. the expression is independent of the order of operational factors.

 

We will explain the method with the help of following

Example: Solve d2y/ dx2 – 5 dy/dx +6y = e3x.

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