Question

Find the solution of the equation:
$dy over dx space equals space straight e to the power of straight x plus straight y end exponent plus straight x squared space straight e to the power of straight x$
subject to the condition, when x = 0; y = 0.

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Solution

dy over dx space equals space straight e to the power of straight x plus straight y end exponent plus straight x squared straight e to the power of straight x space space space space space space rightwards double arrow space space space dy over dx space equals space straight e to the power of straight x. space straight e to the power of straight y space plus space straight x squared space straight e to the power of straight x space space rightwards double arrow space space dy over dx space equals space left parenthesis straight e to the power of straight x plus straight x squared right parenthesis space straight e to the power of straight y

Separating the variables, we get,

dy over straight e to the power of straight y space equals space left parenthesis straight e to the power of straight x plus straight x squared right parenthesis space dx

therefore space space space space space space space integral straight e to the power of negative straight y end exponent dy space equals space integral left parenthesis straight e to the power of straight x plus straight x squared right parenthesis space dx space space space space space rightwards double arrow space space space fraction numerator straight e to the power of negative straight y end exponent over denominator negative 1 end fraction space equals space straight e to the power of straight x plus straight x cubed over 3 plus straight c

straight e to the power of negative straight y end exponent plus straight e to the power of straight x plus straight x cubed over 3 plus straight c space equals space 0

...(1)
When x = 0, y = 0, we have

straight e to the power of 0 plus straight e to the power of 0 plus 0 plus straight c space equals space 0 space space space space rightwards double arrow space space space space 1 plus 1 plus straight c space equals space 0 space space space rightwards double arrow space space space straight c space equals negative space 2

Putting c = -2 in (1), the required solution is

e to the power of italic minus x end exponent italic plus e to the power of x italic plus x to the power of italic 3 over italic 3 italic minus italic 2 italic space italic equals italic space italic 0 italic.

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