Question

How to you find the general solution of `yy'=sinx`?

Answer 1

`y^2=-2cosx+C_1`

Explanation:

`yy'=sinx`

rewite in differential form

`y(dy)/(dx)=sinx`

separate variables and integrate

`intydy=intsinxdx`

`1/2y^2=-cosx+C`

`y^2=-2cosx+C_1`