Find the particular solution of the differential equation
$dy over dx equals negative fraction numerator x plus y space c o s space x over denominator 1 plus s i n space x space end fraction space g i v e n space t h a t space y equals 1 space w h e n space x equals 0$

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Solution

dy over dx equals negative fraction numerator straight x plus ycosx over denominator 1 plus sin space straight x end fraction rightwards double arrow space dy over dx plus fraction numerator cosx over denominator 1 plus sinx end fraction straight y equals negative fraction numerator straight x over denominator 1 plus sin space straight x end fraction space..... space left parenthesis straight i right parenthesis This space is space straight a space linear space differential space equation space with straight P space equals space fraction numerator cosx over denominator 1 plus sin space straight x end fraction comma space straight Q space space equals negative fraction numerator straight x over denominator 1 plus sin space straight x end fraction therefore comma straight I. straight F. space equals space straight e to the power of integral fraction numerator cosx over denominator 1 plus sin space straight x end fraction dx end exponent equals space space straight e to the power of log space left parenthesis 1 plus sinx right parenthesis end exponent equals 1 space plus space sin space straight x Multiplying space both space sides space of space left parenthesis straight i right parenthesis space by space straight I. straight F. space equals space 1 space plus space sin space straight x comma space we space get left parenthesis 1 plus sinx right parenthesis dy over dx plus straight y space cos space straight x space equals space minus straight x Integrating space space with space respect space to space straight x comma space we space get straight y left parenthesis 1 plus sin right parenthesis space equals integral negative xdx space plus space straight C rightwards double arrow space space straight y space equals space fraction numerator 2 straight C minus straight x squared over denominator 2 left parenthesis 1 plus sinx right parenthesis end fraction space...... space left parenthesis ii right parenthesis Given space that space straight y space equals 1 space when space straight x equals space 0 thus comma 1 space equals fraction numerator 2 straight C over denominator 2 left parenthesis 1 plus 0 right parenthesis end fraction rightwards double arrow space space straight C space equals 1 space space....... space left parenthesis iii right parenthesis Put space left parenthesis iii right parenthesis space space in space space left parenthesis ii right parenthesis comma space space we space get straight y space equals space fraction numerator 2 minus straight x squared over denominator 2 left parenthesis 1 plus sin space straight x space right parenthesis end fraction

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Vector Algebra

Find the particular solution of the differential equation
$dy over dx equals negative fraction numerator x plus y space c o s space x over denominator 1 plus s i n space x space end fraction space g i v e n space t h a t space y equals 1 space w h e n space x equals 0$

Some More Questions From Vector Algebra Chapter

Determine order and degree (if defined) of differential equations:
y' + y = e^x

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NCERT Sample Papers

Entrance Exams Preparation

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Vector Algebra

Find the particular solution of the differential equation

Some More Questions From Vector Algebra Chapter

Determine order and degree (if defined) of differential equations:y' + y = ex

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Find the particular solution of the differential equation
$dy over dx equals negative fraction numerator x plus y space c o s space x over denominator 1 plus s i n space x space end fraction space g i v e n space t h a t space y equals 1 space w h e n space x equals 0$

Determine order and degree (if defined) of differential equations:
y' + y = e^x