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Differential Equations

Question

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For the given differential equation, find a particular solution satisfying the given condition:
$dy over dx minus 3 space straight y space cotx space equals space sin space 2 straight x space colon space space straight y space equals space 2 space space when space space straight x space equals space straight pi over 2$

Solution

The given differential equation is

dy over dx minus 3 straight y space cotx space equals space sin space 2 straight x

Comparing it with

dy over dx plus Py space equals straight Q comma space we space get comma space space straight P space equals space minus 3 space cotx space comma space space straight Q space equals space sin space 2 straight x

integral straight P space dx space equals space minus 3 integral cot space straight x space dx space equals space minus 3 space log space sinx

therefore space space straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of negative 3 space log space sinx end exponent space equals space straight e to the power of log space left parenthesis sinx right parenthesis cubed end exponent space equals space left parenthesis sin space straight x right parenthesis to the power of negative 3 end exponent space equals fraction numerator 1 over denominator sin cubed straight x end fraction

therefore

solution of given equation is

straight y. space straight e to the power of integral straight P space dx end exponent space equals space integral straight Q. space straight e to the power of integral straight P space dx end exponent dx plus straight c

or

straight y. space fraction numerator 1 over denominator sin space cubed straight x end fraction space equals space integral sin space 2 straight x. space 1 over sin to the power of 3 straight x end exponent dx plus space straight c

or

fraction numerator straight y over denominator sin cubed straight x end fraction space equals space 2 space integral fraction numerator cosx over denominator sin squared straight x end fraction dx plus straight c

or

fraction numerator straight y over denominator sin cubed straight x end fraction space equals space 2 space integral left parenthesis sinx right parenthesis to the power of negative 2 end exponent space cosx space dx space plus space straight c

or

fraction numerator straight y over denominator sin cubed straight x end fraction space equals 2 fraction numerator left parenthesis sin space straight x right parenthesis to the power of negative 1 end exponent over denominator negative 1 end fraction plus straight c

or

fraction numerator straight y over denominator sin cubed straight x end fraction space equals space minus fraction numerator 2 over denominator sin space straight x end fraction plus straight c

or

straight y space equals space minus 2 space sin squared straight x plus space straight c space sin cubed straight x

...(1)
Initially, y = 2 when

straight x equals space straight pi over 2

therefore

2 space equals space minus 2 sin squared straight pi over 2 plus straight c space sin cubed. space straight pi over 2

rightwards double arrow

2 space equals space minus 2 plus straight c

open square brackets because space space sin space straight pi over 2 space equals space 1 close square brackets

rightwards double arrow

c = 4
Putting c = 4 in (1), we get,

straight y space equals negative 2 space sin squared straight x plus 4 space sin cubed straight x comma space which space is space required space solution. space

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