Question

What is a solution to the differential equation `dy/dx =( 2csc^2 x)/ (cot x)`?

Answer 1

`y = 2ln|sin(x)| - 2ln|cos(x)| + C`

Explanation:

The equation simplifies:

`dy/dx = 2csc(x)sec(x)`

Separate variables:

`dy = 2csc(x)sec(x)dx`

Integrate:

`intdy = 2intcsc(x)sec(x)dx`

`y = 2ln|sin(x)| - 2ln|cos(x)| + C`

Answer 2

`y=2lnabstanx+C`

Explanation:

First we should separate the variables, which means that we can treat `dy/dx` like division. We can move the `dx` to the right hand side of the equation to be with all the other terms including `x`.

`dy=(2csc^2x)/cotxdx`

Now integrate both sides:

`intdy=2intcsc^2x/cotxdx`

On the right hand side, let `u=cotx`. This implies that `du=-csc^2xdx`.

`y=-2int(-csc^2x)/cotxdx`

`y=-2int(du)/u`

`y=-2lnabsu+C`

`y=-2lnabscotx+C`

One possible simplification we could make if we wanted would be to bring the `-1` outside the logarithm into the logarithm as a `-1` power:

`y=2lnabstanx+C`