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The solution of the differential equation (2x*y⁴*e^y + 2x*y³ + y) dx + (x²*y⁴*e^y - x²*y² - 3x) dy = 0 is:
- x²*e^y + x²/y + x/y³ = c
- x²*e^y - x²/y + x/y³ = c
- x²*e^y + x²/y - x/y³ = c
- x²*e^y - x²/y - x/y³ = c
Correct answer: x²*e^y + x²/y + x/y³ = c
Solution
Dividing the original equation by y⁴ gives (2x*e^y + 2x/y + 1/y³)dx + (x²*e^y - x²/y² - 3x/y⁴)dy = 0. This is the total differential of F = x²*e^y + x²/y + x/y³, since dF/dx = 2x*e^y + 2x/y + 1/y³ and dF/dy = x²*e^y - x²/y² - 3x/y⁴. So the solution is x²*e^y + x²/y + x/y³ = c.
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