Question

1. Complementary function of:
day
+62 +9y = 2e-3x
dy
dx
dx2
is :
(A) (C + C2x) e-38
(B) Cax e-3x
(C) (C + C2) et
(D) C2xe-*​

Answer 1

Answer:

Solution :

y=3e2x+2e3x

⇒dydx=3e2x(2)+2e3x(3)

⇒dydx=6(e2x+e3x)

⇒d2ydx2=6(2e2x+3e3x)

Now,L.H.S.=d2ydx2−5dydx+6y

=12e2x+18e3x−30e2x−30e3x+18e2x+12e3x

=0=R.H.S.