Need to find the integral of e^[t^(2) - t]dt
Hello,
This cannot be expressed in terms of simple functions.
Integrate[E^(x^2 - x), x] ==
(Sqrt[Pi]*Erfi[-1/2 + x])/(2*E^(1/4))
http://reference.wolfram.com/mathematica/ref/Erfi.html
thanks. hmm. i probably did something wrong. here is the problem i am working on. we are supposed to use an integrating factor for these linear first order differential equations.
dP/dt + 2tP = P + 4t -2
i manipulated this into "standard form" ( dy/dx + P(x)y = f(x) )
dP/dt + (2t - 1)P = 4t - 2
then the integrating factor would be e^integral of [2t -1]dt, and that would be e^[t^(2) - t] correct?
now my differential equation looks like this
e^[t^(2) - t]y = integral of [ (4t-2)e^[t^(2) - t] ]dt
i separated the right side into
4 integral of te^[t^(2) - t]dt -2 integral of e^[t^(2) - t]dt
now im stuck on integrating that.
Yes, that is correct.
Don't separate! Let u= t^2- t. The integral becomes $\displaystyle 2\int e^u du$now my differential equation looks like this
e^[t^(2) - t]y = integral of [ (4t-2)e^[t^(2) - t] ]dt
i separated the right side into
4 integral of te^[t^(2) - t]dt -2 integral of e^[t^(2) - t]dt
now im stuck on integrating that.