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Sidana [21]
1 year ago
14

2. Find the solution of each differential equation. (a) y/2y-8y = 0 (b) 25y/- 20y + 4y = 0 (c) y + 2y + 2y = 0 2. Find the solut

ion of each differential equation . ( a ) y / 2y - 8y = 0 ( b ) 25y / - 20y + 4y = 0 ( c ) y + 2y + 2y = 0​

Advanced Placement (AP)
1 answer:
jek_recluse [69]1 year ago
3 0

Each of these ODEs is linear and homogeneous with constant coefficients, so we only need to find the roots to their respective characteristic equations.

(a) The characteristic equation for

y'' - 2y' - 8y = 0

is

r^2 - 2r - 8 = (r - 4) (r + 2) = 0

which arises from the ansatz y = e^{rx}.

The characteristic roots are r=4 and r=-2. Then the general solution is

\boxed{y = C_1 e^{4x} + C_2 e^{-2x}}

where C_1,C_2 are arbitrary constants.

(b) The characteristic equation here is

25r^2 - 20r + 4 = (5r - 2)^2 = 0

with a root at r=\frac25 of multiplicity 2. Then the general solution is

\boxed{y = C_1 e^{2/5\,x} + C_2 x e^{2/5\,x}}

(c) The characteristic equation is

r^2 + 2r + 2 = (r + 1)^2 + 1 = 0

with roots at r = -1 \pm i, where i=\sqrt{-1}. Then the general solution is

y = C_1 e^{(-1+i)x} + C_2 e^{(-1-i)x}

Recall Euler's identity,

e^{ix} = \cos(x) + i \sin(x)

Then we can rewrite the solution as

y = C_1 e^{-x} (\cos(x) + i \sin(x)) + C_2 e^{-x} (\cos(x) - i \sin(x))

or even more simply as

\boxed{y = C_1 e^{-x} \cos(x) + C_2 e^{-x} \sin(x)}

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