Question

can anyone help me to solve this bernoulli's equation or substitution..ill give medal

Answer 1

\[(3tanx-2cosy)\sec^2xdx+tanxsinydy=0\]

Answer 2

Can you finish it?

Answer 3

ill try

Answer 4

its hard to analyze that thing lol

Answer 5

@SolomonZelman

Answer 6

If you're not familiar with the method of exact equations, I think the substitution you need here is \(z=\cos y\), which gives \(\mathrm dz=-\sin y\,\mathrm dy\). This changes the ODE to
\[(3\tan x-2z)\sec^2x\,\mathrm dx-\tan x\,\mathrm dz=0\]Another substitution of \(u=\tan x\) gives \(\mathrm du=\sec^2x\,\mathrm dx\), so you now have
\[(3u-2z)\,\mathrm du-u\,\mathrm dz=0\iff u\frac{\mathrm dz}{\mathrm du}+2z=3u\]which is linear.