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Differential Equations

Question
CBSEENMA12033211

Find a particular solution of the differential equation left parenthesis straight x plus 1 right parenthesis space dy over dx space equals space 2 straight e to the power of negative straight y end exponent minus 1 comma space space given that y = 0,  when x = 0.

Solution
The given differential equation is
                   left parenthesis straight x plus 1 right parenthesis space straight y apostrophe space space equals space 2 space straight e to the power of negative straight y end exponent minus 1 space space space space or space space left parenthesis straight x plus 1 right parenthesis space dy over dx space equals space 2 straight e to the power of negative straight y end exponent minus 1
or           dy over dx space equals space fraction numerator 2 over denominator straight x plus 1 end fraction space straight e to the power of negative straight y end exponent minus fraction numerator 1 over denominator straight x plus 1 end fraction
or      straight e to the power of straight y dy over dx plus fraction numerator 1 over denominator straight x plus 1 end fraction straight e to the power of straight y space equals space fraction numerator 2 over denominator straight x plus 1 end fraction
Put straight e to the power of straight y space equals space straight z comma space space space space space therefore space space straight e to the power of straight y dy over dx space equals space dz over dx
therefore        dz over dx plus fraction numerator 1 over denominator straight x plus 1 end fraction straight z space equals fraction numerator 2 over denominator straight x plus 1 end fraction
Comparing it with dz over dx plus Pz space equals space straight Q comma space space we space get comma
                straight P space equals fraction numerator 1 over denominator straight x plus 1 end fraction. space space space straight Q space space equals space fraction numerator 2 over denominator straight x plus 1 end fraction
            straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of integral fraction numerator 1 over denominator straight x plus 1 end fraction end exponent space equals space straight e to the power of log left parenthesis straight x plus 1 right parenthesis end exponent space equals space straight x plus 1
therefore  solution of differential equation is
                straight z space. space straight e to the power of integral straight P space dx end exponent space equals space integral space straight Q. space straight e to the power of integral straight P space dx end exponent space plus straight c space
or              straight z left parenthesis straight x plus 1 right parenthesis space equals space integral open parentheses fraction numerator 2 over denominator straight x plus 1 end fraction close parentheses space left parenthesis straight x plus 1 right parenthesis space dx plus straight c
or space space space space space space straight z left parenthesis straight x plus 1 right parenthesis space equals space integral 2 dx plus straight c
or space space space space space straight z left parenthesis straight x plus 1 right parenthesis space equals space 2 straight x plus straight c
or space space space straight e to the power of straight y left parenthesis straight x plus 1 right parenthesis space equals space 2 straight x plus straight c 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 space space space space space space space space... left parenthesis 1 right parenthesis
Now y = 0    when   x = 0
therefore space space space space space space straight e to the power of 0 left parenthesis 0 plus 1 right parenthesis space equals space 0 plus straight c space space space space space space space space space space space rightwards double arrow space space space straight c space equals space 1
therefore space space space from space space left parenthesis 1 right parenthesis comma space space space straight e to the power of straight y space left parenthesis straight x plus 1 right parenthesis space equals space 2 straight x plus 1
which is required solution.
           
                            
        

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