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Differential Equations

Question
CBSEENMA12033191

Solve:
dy over dx plus 3 straight y space equals space sin space 2 straight x

Solution
The given differential equation is
                         dy over dx plus 3 straight y space equals space sin space 2 straight x
Comparing it with dy over dx plus straight P space straight y space equals space straight Q comma space we space get comma space
                    straight P space equals space 3 comma space space space straight Q space equals sin space 2 straight x
integral straight P space dx space equals space integral 3 space dx space equals space 3 straight x
           straight I. straight F. space equals straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of 3 straight x end exponent
Solution of given differential equation is
                    ye 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 straight e to the power of 3 straight x end exponent space equals space integral straight e to the power of 3 straight x end exponent space space sin space 2 straight x space dx space plus straight c
or     ye to the power of 3 straight x end exponent space equals space fraction numerator 1 over denominator square root of 9 plus 4 end root end fraction straight e to the power of 3 straight x end exponent space sin space open parentheses 2 straight x minus tan to the power of negative 1 end exponent 2 over 3 close parentheses plus straight c
                                             open square brackets because space space space integral straight e to the power of ax space sin space bx space dx equals space fraction numerator 1 over denominator square root of straight a squared plus straight b squared end root end fraction straight e to the power of ax space sin space open parentheses bx minus tan to the power of negative 1 end exponent straight b over straight a close parentheses close square brackets
therefore space space space space space straight y space equals space fraction numerator 1 over denominator square root of 13 end fraction sin space open parentheses 2 straight x minus tan to the power of negative 1 end exponent 2 over 3 close parentheses plus space straight c space straight e to the power of negative 3 straight x end exponent
which is required solution. 

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