Question

How do you find the solution to the differential equation `dy/dx=cos(x)/y^2` where `y(π/2)=0`?

Answer 1

You can write:
`y^2dy=cos(x)dx`
and integrate;
`inty^2dy=intcos(x)dx`
Which gives you:
`y^3/3=sin(x)+c`
`y^3=3sin(x)+c`
Now we find the value of `c`;
`(0)^3=3sin(pi/2)+c`
`0=3+c`
And `c=-3`
your particular solution is then:
`y^3=3sin(x)-3`