Weird Question

dy/dx = y.x^2.cosx
1/y dy = x^2.cos x dx

ln|y| = x^2.sinx - int (2x.sinx) dx = x^2.sinx - (2x(-cosx) - int (-2cosx))dx

ln|y| = x^2.sinx + 2x.cosx - 2sinx + C

Oooooh, so it's a question in the form:

dy/dx = f(x).g(x) which implies that:

integral 1/g(y).dy = integral f(x).dx + c

Silly me, I think I need to read through all of my notes again this weekend, before I try and do hard questions, esp those on paper we got for hw! What do I do then about the values of y and x it initially gives me? Do I put them into equation and solve to get a value of c?

Yea, sub them in to find c = 2pi.

Good

Sometimes you're asked to give write it in the form y = something, in which case you put e^everything, since e^lny = y.