# Weird Question

Found interesting question on P3 mock paper that I'm not sure aout:-

Given that y=1 at x=pi, solve the differential equation:

dy/dx = y.x^2.cosx (y>0)

Any ideas? Never seen a question of this style before.

I thought that dy/dx had originated from 2 parametric equations (with t=parameter) and maybe dy/dt = y and dx/dt=secx/x^2?! Then thought about integrating them to get two equations?! Not sure though, any ideas!? Thanks! I'll probably end up posting a vector question as well tomorrow that I can't do, but I'm determined to work it out for myelf!
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
JamesF
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?
Hoofbeat
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.
JamesF
Yea, sub them in to find c = 2pi.

Cheers! So then I just leave it in the form lny=...bla blah?!

Wickd! Having read through how you did your method I then attempted the question on my own and I got same as you, so v.pleased :d
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.
JamesF
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.

James, you're a superb mathematician for your age.