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\[y^3+\cos(t)y'=2+y \sin(t)\]is this the question?

does not look exact though without an integration factor.

\[y^3-y \sin(t)-2+\cos(t)y'=0\]

the y(0)=-1 is just an initial condition, the test for exactness is applied before that:
\[M_y=N_x\]

not quite, and
\[f_y=(y^3+\cos(t))\]

without the y'

yes, what will you get from that?

yeh you got it almost.

very good, but also check for the cos(t) term, it's just a constant, so you add a y infront of it.

yes because your dimension/direction of integration is dy, so t is constant.

yes, how would you continue?

yes, integrate that.

almost it would give you
ycos(t)-2t+g(y)

yes you still need to find C, do that with your initial values.

I believe H(t,y(0))=-2t+C=-1

maybe you want to copy both answers to your problem sheet though, whichever your Professors prefer.

you're done

no there is only one constant of integration

you're welcome