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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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