Maclaurin series for y' = (1 - xy)^(1/2), up to x^3

G

Guest

Guest
Find Maclaurin series up to \(\displaystyle x^3\)

y' = (1 - xy)^(1/2)

Is this a kind of differential equation, or what? How would I solve it?
 
Were Maclaurin series not covered in your class or text?

Thank you.

Eliz.
 
i don't have class or text
i'm a self learner

please tell me how to solve it
 
I'm sorry, but we really can't teach courses within this environment. For self-study teaching assistance, I would suggest you hire a tutor local to your area, who can work with you face-to-face to help you learn the material. A few hours a week should be sufficient.

Good luck! :D

Eliz.
 
I ONLY KNOW that Maclaurin's expansion =

f(x) = f(0)+xf'(0)+\(\displaystyle \frac{x^2}{2!}f''(0)+.......+\frac{x^n}{n!}f^{(n)}\)


then what should I do with the equation ??
Should I solve the equation first to get y = .....?

Thanks a lot. :)
 
You must log in or register to reply here.
Top