I am given the equation:
$y^3 + y^2 + y - 3 = x$
with $y(0)=1$. I am wondering if what I have done is valid, given that this is a homework question for an Integral Calculus class, but I seem to be able to answer the question without doing any integration.
Taking the derivative of both sides (and using implicit differentiation) with respect to $x$ and then rearranging, I have
$\frac{d}{dx} \left(y^3 + y^2 + y - 3\right) = \frac{d}{dx}(x)$
$\frac{dy}{dx} = \frac{1}{3y^2 + 2y + 1}$
So, since I am given $y(0) = 1$, I have my first two terms in the Taylor Series: f(0) = 1, and now f'(0) = 1/6, since y = 1 when x = 0.
I then take the 2nd derivative with respect to x (and again using implicit differentiation) to get:
$d^2y/dx^2 = - y'(x)(6y+2)/(3y^2 + 2y + 1)^2$
Plugging in for y and y', when x = 0, and I get y''(0) = -1/27
Therefore my Taylor polynomial of order 2, about x = 0, is:
$1 + x/6 - x^2/27$
Is this correct?
I appreciate any tips and advice!