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Why was the calculus student confused about y equals ex and the derivative of y equals ex?
Because the derivative of e^x is e^x (the original function back again). This is the only function that has this behavior.
Y' equals x plus y Explicit solution?
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.
What is the derivative of e-x?
I suppose you mean ex. The derivate is also ex.
How do you solve for x where e2x plus ex equals 0?
e2x + ex = 0e2x = -exe2x / ex = -1ex = -1x = ln(-1)Actually, that's just a restatement of the original problem, but that'sas 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.
How do you differentiate exp-ax2?
I'm not sure if you are asking about: e-ax2 or e-ax^2. Use the carets=^ in the future to raise something to a power.For e-ax2:You need to remember that: (ex)`= exalso: (eax)`= aeax, assuming a is a constant and not a function of x.Justrecognizeany constant drops down to the front.Therefore: (e-ax2)`=-2ae-ax2, treat the negative just as a constant or (-1)For e-ax^2: this one is a little bit trickier:We can just use u-substitution.Let u = x2.u` = 2xIf we substitute the u for x^2, we have: e-au , and we know the deriviate of that is -ae-2au, but we need to multiply by u` as well to get:(-ae-2au)(2x) = -2xae-2au, then we can just plug u back in.u = x2, so:-2xae-2au = -2xae-2ax^2And that is the answer.(e-ax^2)`= -2xae-2ax^2
