Question

Find one-parameter families of solution curves of the following differential equation:
$straight y apostrophe plus straight y space cosx space equals space straight e to the power of sin space straight x end exponent space cos space straight x$

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Solution

The given differential equation is

straight y apostrophe plus straight y space cosx space equals space straight e to the power of sinx space cos space straight x space space space space space space space or space space space space space dy over dx plus left parenthesis cos space straight x right parenthesis. space straight y space equals space straight e to the power of sin space straight x end exponent space space cosx space

Comparing it with

dy over dx plus straight P space straight y space equals space straight Q comma space we space get space straight P space equals cosx comma space straight Q space equals space straight e to the power of sinx. space cosx

therefore space space space space space space space space straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of integral space cosx space dx end exponent space equals straight e to the power of sin space straight x end exponent

Solution of given differential 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

straight y space straight e to the power of sinx space equals space integral straight e to the power of sinx space cosx. space straight e to the power of sinx dx plus straight c

straight y space straight e to the power of sinx space equals space integral straight e to the power of 2 space sinx end exponent space cosx space dx plus straight c

...(1)
Let I =

integral straight e to the power of 2 sinx end exponent space cos space straight x space dx

Put sinx = t,

therefore space space space cos space straight x space dx space equals dt

therefore

I =

integral straight e to the power of 2 straight t end exponent dt space equals space straight e to the power of 2 straight t end exponent over 2 space equals 1 half straight e to the power of 2 sinx end exponent

therefore space space from space left parenthesis 1 right parenthesis comma space space space ye to the power of sinx space equals space 1 half straight e to the power of 2 sinx end exponent plus straight c comma space which space is space required space solution.

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