constants can be brought outside the integration symbol as a constant multiplier. for example: S12dx is the same as 12Sdx which is the same as 12x+C however: S(x+3)dx is Sxdx+S3dx not 3Sxdx
Chat with our AI personalities
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.
I suppose you mean ex. The derivate is also ex.
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.
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