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प्रश्न
उत्तर
`("d"y)/("d"x)` = e(x+y) + x2ey
∴ `("d"y)/("d"x)` = ex . ey + x2 ey
∴ `("d"y)/("d"x)` = ey(ex + x2)
∴ `("d"y)/("e"^y)` = (ex + x2) dx
Integrating on both sides, we get
`int_"e"^(-y) "d"y = int("e"^x + x^2) "d"x`
∴ `("e"^(-y))/(-1) = "e"^x + x^3/3 + "c"_1`
∴ e−y = `- "e"^x - x^3/3 - "c"_1`
∴ `"e"^(-y) + "e"^x + x^3/3` = c, where c = c1
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