Math, asked by neha16403, 8 months ago

Solve the differential equation :-
(cosx.cosy - cotx) dx - (sinx.siny) dy=0​

Answers

Answered by pulakmath007
12

\displaystyle\huge\red{\underline{\underline{Solution}}}

TO CALCULATE

The solution of the Differential equation

 \sf{  (\cos x \cos y -  \cot x)dx -  \sin x \sin y \: dy\: } = 0

FORMULA TO BE IMPLEMENTED

  \sf{1. \: d(uv) = v \: du  + u \: dv\:  \: }

CALCULATION

The given differential equation is

 \sf{  (\cos x \cos y -  \cot x)dx -  \sin x \sin y \: dy\: } = 0

Rearranging we get

 \implies  \sf{  \cos x \cos y \: dx -  \sin x \sin y \: dy\: - \cot xdx } = 0

 \implies  \sf{  \cos y \: d( \sin x)   +  \sin x \: d( \cos y )\: \: - \cot x \: dx } = 0

 \implies  \sf{  d \: (  \sin x \:  \cos y )\: \: - \cot x \: dx } = 0

On integration

 \displaystyle \:   \sf{  \int d \: (  \sin x \:  \cos y )\: \: -  \int\cot x \: dx } = 0 \:  \:  \:

  \implies \: \displaystyle \:   \sf{  \int d \: (  \sin x \:  \cos y )\: \: -  \int \frac{ \cos x}{ \sin x}   \: dx } = 0 \:  \:  \: ....(1)

 \sf{Let \:   \: \:z =  \sin x }

 \sf{dz =  \cos x \: dx}

Equation (1) becomes

  \implies \: \displaystyle \:   \sf{  \int d \: (  \sin x \:  \cos y )\: \: -  \int \frac{dz}{ z}    } = 0 \:  \:  \:

  \implies \: \displaystyle \:   \sf{   (  \sin x \:  \cos y )\: \: -  \ln |z|     } = c \:  \:  \:

Where C is integration constant

  \implies \: \displaystyle \:   \sf{   (  \sin x \:  \cos y )\: \: -  \ln | \sin x \: |     } = c \:  \:  \:

Which is the required solution of the Differential equation

RESULT

The solution is

 \boxed{ \:  \: \displaystyle \:   \sf{   (  \sin x \:  \cos y )\: \: -  \ln | \sin x \: |     } = c \:  \:  \: \: }

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