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

Answer 1

Paisley Johnson

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

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#

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Answer 2

Isabella Ackerman

#y=2lnabstanx+C#

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#

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Answer 3

Evelyn Agard

The solution to the differential equation ( \frac{dy}{dx} = \frac{2 \csc^2 x}{\cot x} ) is ( y = -2 \cot x + C ), where ( C ) is the constant of integration.

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Answer from HIX Tutor

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