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

is

which arises from the ansatz
.
The characteristic roots are
and
. Then the general solution is

where
are arbitrary constants.
(b) The characteristic equation here is

with a root at
of multiplicity 2. Then the general solution is

(c) The characteristic equation is

with roots at
, where
. Then the general solution is

Recall Euler's identity,

Then we can rewrite the solution as

or even more simply as

Answer:
The magnetic field of the earth was discovered by Carl Friedrich Gauss in 1835.
Of course it's possible, because it means that sintheta=-2/5 which satisfies this -1<=sintheta<=1
Is it “b” ? Please tell me please
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