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The general solution of the differential equation xxdy+(yex+2x) dx = 0 -

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Question

The general solution of the differential equation `x^xdy + (ye^x + 2x)  dx` = 0

Options

  • `xe^y + x^2 = c`

  • `xe^y + y^2 = c`

  • `ye^x + x^2 = c`

  • `ye^y + x^2 = c`

MCQ

Solution

`ye^x + x^2 = c`

Explanation:

The differential equation is `e^x  (dy)/(dx) + e^xy = - 2x` 

or `(dy)/(dx) + 1  * y = - 2xe^-x`

I.F. = `e^(int dx = e^x)`

∴ Solution is `ye^x = int (- 2x)e^-x xx e^x  dx + c`

= `- int 2x  dx + c = - x^2 + c`

⇒ `ye^x + x^2 = c`

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