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Yuki888 [10]
2 years ago
5

I need to get the solution

Advanced Placement (AP)
1 answer:
katen-ka-za [31]2 years ago
4 0

yeah im sorry but i dont know

:c

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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

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

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(b) The characteristic equation here is

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\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

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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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