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

2 Answers
Dec 8, 2016

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

Dec 8, 2016

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