a) The limit of (y+1)/sin(x) as x→0 is of the indeterminate form 0/0, so you need to use l'Hospital's rule:

lim_{x→0}(y+1)/sin(x) = lim_{x→0} y'/cos(x) = lim y²(2x+2)/cos(x) = 2.

b) Euler's method tells us that y_{n+1}=y_{n} + h y'(x_{n},y_{n}).

If you go in two steps of equal size from 0 to 1/2, the step size h must equal 1/4:

y_{n+1}= y_{n} + (1/4) y'(x_{n},y_{n})

y_{1}= y_{0} + (1/4) y'(x_{0},y_{0}) = -1+(1/4) 2 = -1/2 ≈ y(1/4)

y_{2}= y_{1} + (1/4) y'(x_{1},y_{1}) = -1/2 + (1/4)(5/8) = -11/32 ≈ y(1/2).

c) The equation is separable:

dy/y² = (2x+2) dx ⇒ -1/y = x²+2x+c ⇒ y = -1/(x²+2x+c)

With y(0)=-1 we get c=1, so that

y = -1/(x²+2x+1).

Note that y(1/2) = -4/9.

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