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Vector Algebra

Question
CBSEENMA12033071
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Find the particular solution of the differential equation
(1 + e2x ) dy + (1 + y2 ) ex dx = 0, given that y = 1 when x = 0.

Solution

The given differential equation is
              (1 + e2x ) dy + (1 + y2 ) ex dx = 0   or   (1 + e2x ) dy  = - (1 + y2 ) ex dx
therefore space space space space fraction numerator 1 over denominator 1 plus straight y squared end fraction dy space equals negative space fraction numerator straight e to the power of straight x over denominator 1 plus straight e to the power of 2 straight x end exponent end fraction dx
rightwards double arrow space space space space integral fraction numerator 1 over denominator 1 plus straight y squared end fraction dy space equals space minus integral fraction numerator straight e to the power of straight x over denominator 1 plus straight e to the power of 2 straight x end exponent end fraction dx 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
Let I = integral fraction numerator straight e to the power of straight x over denominator 1 plus straight e to the power of 2 straight x end exponent end fraction dx
Put         straight e to the power of straight x space equals space straight t comma space space space therefore space space space space straight e to the power of straight x space dx space equals space dt
therefore space space space space straight I space equals space integral fraction numerator dt over denominator 1 plus straight t squared end fraction space equals space tan to the power of negative 1 end exponent straight t space equals space tan to the power of negative 1 end exponent left parenthesis straight e to the power of straight x right parenthesis therefore space space space space from space left parenthesis 2 right parenthesis comma space space tan to the power of negative 1 end exponent straight y space equals space tan left parenthesis straight e to the power of straight x right parenthesis space 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... left parenthesis 2 right parenthesis
Now,   x = 0,  y = 1
therefore space space space tan to the power of negative 1 end exponent 1 space equals space minus tan left parenthesis straight e to the power of 0 right parenthesis space plus space straight c space space space rightwards double arrow space space space space straight pi over 4 space equals space minus tan to the power of negative 1 end exponent left parenthesis 1 right parenthesis space plus space straight c rightwards double arrow space space space straight pi over 4 space equals space minus straight pi over 4 plus straight c space space space rightwards double arrow space space space space straight c space equals space straight pi over 2 therefore space space from space left parenthesis 2 right parenthesis comma space tan to the power of negative 1 end exponent straight y equals space minus tan space straight e to the power of straight x plus straight pi over 2 comma space which space is space required space solution. space

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