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

Dec 8, 2016

$y = 2 \ln | \sin \left(x\right) | - 2 \ln | \cos \left(x\right) | + C$

Explanation:

The equation simplifies:

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 \csc \left(x\right) \sec \left(x\right)$

Separate variables:

$\mathrm{dy} = 2 \csc \left(x\right) \sec \left(x\right) \mathrm{dx}$

Integrate:

$\int \mathrm{dy} = 2 \int \csc \left(x\right) \sec \left(x\right) \mathrm{dx}$

$y = 2 \ln | \sin \left(x\right) | - 2 \ln | \cos \left(x\right) | + C$

Dec 8, 2016

$y = 2 \ln \left\mid \tan \right\mid x + C$

Explanation:

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

$\mathrm{dy} = \frac{2 {\csc}^{2} x}{\cot} x \mathrm{dx}$

Now integrate both sides:

$\int \mathrm{dy} = 2 \int {\csc}^{2} \frac{x}{\cot} x \mathrm{dx}$

On the right hand side, let $u = \cot x$. This implies that $\mathrm{du} = - {\csc}^{2} x \mathrm{dx}$.

$y = - 2 \int \frac{- {\csc}^{2} x}{\cot} x \mathrm{dx}$

$y = - 2 \int \frac{\mathrm{du}}{u}$

$y = - 2 \ln \left\mid u \right\mid + C$

$y = - 2 \ln \left\mid \cot \right\mid x + C$

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:

$y = 2 \ln \left\mid \tan \right\mid x + C$