Solving 4y^4−4x^4+x+y=0 (equation system of partial derivates)

wasangagac4

wasangagac4

Answered question

2022-10-11

Solving 4 y 4 4 x 4 + x + y = 0 (equation system of partial derivates)
Help solving the following equation system:
x = 8 x y + 4 y 2 + y x 2 + y 2 = 0
y = 8 x y + 4 x 2 x x 2 + y 2 = 0

Answer & Explanation

domwaheights0m

domwaheights0m

Beginner2022-10-12Added 11 answers

Multiply each side of the first equation by y, each side of the second equation by x, and add. We get 8 x 2 y + 4 x y 2 + 8 x y 2 + 4 x 2 y = 0, that is, 12 x y ( x + y ) = 0.
From this we get that x = y or x = 0 or y = 0. Substitute in the first equation to finish.
Jaelyn Payne

Jaelyn Payne

Beginner2022-10-13Added 2 answers

Write your system of equations as
8 x y = 4 y 2 + y x 2 + y 2 = 4 x 2 x x 2 + y 2
Then
4 x 2 4 y 2 = 4 ( x + y ) ( x y ) = x + y x 2 + y 2
Since x + y is a factor of both sides, one possibility is x + y = 0. The other is 4 ( x y ) = 1 / ( x 2 + y 2 ), in which case your first equation becomes 8 x y = 4 y 2 + 4 ( x y ) y = 4 x y. But then x = 0 or y = 0.

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