What is a solution to the differential equation dydx=2csc2xcot⁡x?

First we should separate the variables, which means that we can treat dydx 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=2csc2xcot⁡xdxNow integrate both sides:∫dy=2∫csc2xcot⁡xdxOn the right hand side, let u=cot⁡x. This implies that du=−csc2xdxy=−2∫−csc2xcot⁡xdxy=−2∫duuy=−2ln|u|+Cy=−2ln|cot⁡x|+COne possible simplification we could make if we wanted would be to bring the -1outside the logarithm into the logarithm as a-1 power:y=2ln|tan⁡x|+C

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High School
Calculus and Analysis
Calculus 2
Applications of integrals

Breanna Fisher

Answered question

2022-04-17

What is a solution to the differential equation

\frac{d y}{d x} = \frac{2 \csc^{2} x}{\cot x}

Answer & Explanation

ncruuk7ikt

Beginner2022-04-18Added 12 answers

The equation simplifies:

\frac{d y}{d x} = 2 \csc (x) \sec (x)

Separate variables:

d y = 2 \csc (x) \sec (x) d x

Integrate:

\int d y = 2 \int \csc (x) \sec (x) d x

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

regulerenes4w

Beginner2022-04-19Added 10 answers

First we should separate the variables, which means that we can treat

\frac{d y}{d x}

like division. We can move the dx to the right hand side of the equation to be with all the other terms including x.

d y = \frac{2 \csc^{2} x}{\cot x} d x

Now integrate both sides:

\int d y = 2 \int \frac{\csc^{2} x}{\cot x} d x

On the right hand side, let

u = \cot x

. This implies that

d u = - \csc^{2} x d x

y = - 2 \int \frac{- \csc^{2} x}{\cot x} d x

y = - 2 \int \frac{d u}{u}

y = - 2 \ln | u | + C

y = - 2 \ln | \cot 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 | \tan x | + C

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