Question

solve (d^-16)y=0
$(d ^{2} - 16)y = 0$

Answer 1

Solution:

_____________________________________________________________

Given:

(d²-16)y =0,.

_____________________________________________________________

To find:
value of d,
_____________________________________________________________

As it is given that,.

(d²-16)y = 0

d²-16 =0

d² = 16

d = +4 (or) d= -4

_____________________________________________________________


                                    Hope it Helps !!

Answer 2

Answer:

Given (D2 + 16) y =0  The auxiliary equation is m2 + 16 = 0  ⇒ m2 = -16  ⇒ m = ± 4i  It is of the form α ± iβ, α = 0, β = 4  

The complementary function (C.F) is e0x [A cos 4x + B sin 4x]

The general solution is y = [A cos 4x + B sin 4x]

Explanation:

Differential equations are higher math's for school students. But, since this question has come here, we are solving for our curiosity.

Here, D^4 means differentiation of degree four. Let y = f(x); then the given equation is telling that : f’ ’ ’ ’(x) - 16y = 0.

It can be taken as (D^4 - 16) . y = 0, without any mistake.

The D^4 - 16 is called Auxiliary equation that we have to solve.

Now, since y ≠ 0, Then D^4 - 16 = 0,

Hence, D = 2, -2, 2i , -2i any of four roots.

For ±2 ,we have general solution : y = c1 e^{2x} +c2 e^{-2x}

For ±2i ,we have general solution : y = c3 cos(2x) +c4 sin(2x)

So, we have the general solution y =c1 e^{2x} +c2 e^{-2x} + c3 cos(2x) +c4 sin(2x) where, c1,c2,c3 and c4 are any constant (including zero)

solve (d^-16)y=0

https://brainly.in/question/2965338

The complementary function of (D^2-16) y=0 is​

https://brainly.in/question/40984201

#SPJ3