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Find the general solution to y" – 4y' +

8y = 0. Show necessary steps and reasoning that lead to
the answer.

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To find the general solution to the given differential equation:

y" - 4y' + 8y = 0

We can start by assuming a solution of the form y = e^(rt), where r is a constant. Substituting this into the differential equation, we get:

r^2e^(rt) - 4re^(rt) + 8e^(rt) = 0

Factoring out e^(rt), we have:

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

For this equation to hold, either e^(rt) = 0 (which is not possible) or r^2 - 4r + 8 = 0. Solving the quadratic equation, we find the roots:

r = (4 ± √(4^2 - 4 * 1 * 8)) / (2 * 1)

r = (4 ± √(-16)) / 2

r = (4 ± 4i) / 2

r = 2 ± 2i

The roots are complex, so we have:

r1 = 2 + 2i

r2 = 2 - 2i

Since the roots are complex conjugates, we can write the general solution as:

y = c1e^(2t)cos(2t) + c2e^(2t)sin(2t)

where c1 and c2 are arbitrary constants.

Therefore, the general solution to the given differential equation is:

y = c1e^(2t)cos(2t) + c2e^(2t)sin(2t)

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