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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
How do you integrate e?
I assume you mean ex ? If so, by definition: ∫ex dx = ex + C Most calculus textbooks have a table of integrals which will list the integrals of other common forms of exponential & logarithmic functions.
What is the definition of constant used in math?
A number without a variable. Ex.) 5 is a constant, 5x is not.
What is the differentiation and integration?
In Calculus, differentiation is when you apply the theorems to get the derived equation at a given rate, for example you have the velocity function and if you take its derivative, it will give you an acceleration function related to its velocity. Derivatives are often denoted as f'(x) or y'. Integration on the other hand is undoing differentiation. for ex, if you integrate acceleration equation, it will give you a velocity equation.
What is the integral of e to the power of x with respect to x?
∫ ex dx = ex + CC is the constant of integration.
What is an constant term in math?
The y-intercept. Ex: AX-BY=C The 'C' stands for constant and is the y-intercept. Was this helpful?
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.
What is the purpose of differential calculus?
Differential Calculus is to take the derivative of the function. It is important as it can be applied and supports other branches of science. For ex, If you have a velocity function, you can get its acceleration function by taking its derivative, same relationship as well with area and volume formulas.
How do you integrate 2 power 2x?
Keep in mind that the integral of ex = ex and 2x = ex / ln 2. You can then rewrite the exponent as u = 2x, convert dx to du, and from there it's pretty straightforward. (I've left off the "+ C" part, because you should just know that.)
A number that is fixed and does not change in value?
a constant ex: Pi. it will always be 3.14159... it will never change in value.
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 definition of constant-?
Constant means continuing non-stop.to remain in the same place without any movement
A line with a slope of zero?
Has only one variable and one constant and is perpendicular to that variable's axis. Ex) y = 3
