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Differential Equations

Question
CBSEENMA12033217
Wired Faculty App

The general solution of a differential equation of the type dx over dy plus straight P subscript 1 straight x space equals space straight Q subscript 1 is 
  • straight y space straight e to the power of integral straight P subscript 1 dy end exponent space equals space integral space open parentheses straight Q subscript 1 space straight e to the power of integral straight P subscript 1 dy end exponent close parentheses space dy plus straight C
  • straight y. space straight e to the power of integral straight P subscript 1 dx end exponent space equals space integral space open parentheses straight Q subscript 1 space straight e to the power of integral straight P subscript 1 dx end exponent close parentheses dx plus straight C
  • straight x. straight e to the power of integral straight P subscript 1 dy end exponent space equals space integral open parentheses straight Q subscript 1 straight e to the power of integral straight P subscript 1 dy end exponent close parentheses dy plus straight C
  • straight x space straight e to the power of integral straight P subscript 1 dx end exponent space equals space integral open parentheses straight Q subscript 1 space straight e to the power of integral straight P subscript 1 dx end exponent close parentheses dx plus straight C

Solution

C.

straight x. straight e to the power of integral straight P subscript 1 dy end exponent space equals space integral open parentheses straight Q subscript 1 straight e to the power of integral straight P subscript 1 dy end exponent close parentheses dy plus straight C

The given differential equation is dx over dy plus straight P subscript 1 straight x space equals space straight Q subscript 1
Multiplying through by straight e to the power of integral straight P subscript 1 dy end exponent comma space we get,
                     dx over dy straight e to the power of integral straight P subscript 1 dy end exponent space plus space straight P subscript 1 straight x space straight e to the power of integral straight P subscript 1 dy end exponent space equals straight Q subscript 1 space straight e to the power of integral straight P subscript 1 end exponent space dy
or         straight d over dy left parenthesis space straight x space straight e to the power of integral straight P subscript 1 dy end exponent right parenthesis space equals space straight Q subscript 1 space straight e to the power of integral straight P subscript 1 end exponent dy
                              open square brackets because space space straight d over dy left parenthesis straight x space straight e to the power of integral straight P subscript 1 dy end exponent right parenthesis space equals space xe to the power of integral straight P subscript 1 dy. end exponent straight d over dy left parenthesis integral straight P subscript 1 space dy right parenthesis plus straight e to the power of integral straight P subscript 1 dy end exponent dx over dy space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space xe to the power of integral straight P subscript 1 dy end exponent space space straight P subscript 1 space plus space straight e to the power of integral straight P subscript 1 dy end exponent space dx over dy space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space dx over dy straight e to the power of integral straight P subscript 1 dy end exponent space plus space straight P subscript 1 space xe to the power of integral straight P subscript 1 dy end exponent close square brackets space space space
Integrating both sides w.r.t y, we get,
                     xe to the power of integral straight P subscript 1 dy end exponent space equals space integral straight Q subscript 1 space straight e to the power of integral straight P subscript 1 dy end exponent dy plus straight C space which space is space required space solution. space
therefore space space space left parenthesis straight C right parenthesis space is space correct space answer. space

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