e2x + ex = 0
e2x = -ex
e2x / ex = -1
ex = -1
x = ln(-1)
Actually, that's just a restatement of the original problem, but that's
as far as you can go with it, because there's no such thing as the log
(or the ln) of a negative number. Your equation has no solution.
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meant to be e^x = 2e^1-2x
e2x=ex^2 basically means that x2=2x, in which case x2-2x=0, x = 0, 2. I don't think that's what the question meant. It could mean: e2x=(ex)2 . Which comes from one of the rules of exponents. Basically, look at it this way: Take the natural log of both sides: ln e2x= ln(ex)2 From rules of logs: (2x) ln e = (2) ln ex 2x ln e = (2) (x) ln e 2x = (2)(x)
ex = x3 This has two solutions: x = 4.5364... and x = 1.85718... Plot the graph of each and you can see the intersections.
Because the derivative of e^x is e^x (the original function back again). This is the only function that has this behavior.
using Laplace transform, we have: sY(s) = Y(s) + 1/(s2) ---> (s-1)Y(s) = 1/(s2), and Y(s) = 1/[(s2)(s-1)]From the Laplace table, this is ex - x -1, which satisfies the original differential eq.derivative of [ex - x -1] = ex -1; so, ex - 1 = ex - x - 1 + xto account for initial conditions, we need to multiply the ex term by a constant CSo y = C*ex - x - 1, and y' = C*ex - 1, with the constant C, to be determined from the initial conditions.