Question

Solve the differential equation y' = x^2 + x^(-1), given y(-1) = 2.

Answer 1

Pretty much this, then:
\[\Large \frac{dy}{dx} = x^2 + \frac1x\]
?

Answer 2

@creampuffu wake up lol

Answer 3

Yeah, that's what it looks like?

Answer 4

Do I evaluate for the antiderivative then solve for C?

Answer 5

Pretty much, yes :D ^_^\[\Large \int \frac{dy}{dx}dx = \int \left(x^2 + \frac1x\right)dx\]\[\Large \int \frac{dy}{\cancel{dx}}\cancel{dx} = \int \left(x^2 + \frac1x\right)dx\]\[\Large \int dy = \int \left(x^2 + \frac1x\right)dx\]

Answer 6

\[\frac{ x^3 }{ 3 }+lnx + c\] then -1/3 +ln(1) + c = 2?
c=7/3

Answer 7

Yeah, that's right.
You sure that you actually needed help here? XD

By the way, just to avoid those nasty nitpicking instructors, \[\Large \int \frac1x dx = \ln|x| + C\]
don't forget those absolute value bars, actually, or else ln(-1) won't make sense ^_^

But other than that, good job :)