What is the answer to solvefor y,x^4y=(x^8)/8-(x^2)/2+c ?

The answer to solvefor y,x^4y=(x^8)/8-(x^2)/2+c is y=(x^8-4x^2+8c)/(8x^4);\quad x\ne 0

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solvefor y,x^4y=(x^8)/8-(x^2)/2+c

solve for

y, x^{4} y = \frac{x ^{8}}{8} - \frac{x ^{2}}{2} + c

y = \frac{x ^{8} - 4 x ^{2} + 8 c}{8 x ^{4}}; x \neq = 0

Solution steps

x^{4} y = \frac{x ^{8}}{8} - \frac{x ^{2}}{2} + c

Divide both sides by

x^{4}; x \neq = 0

y = \frac{x ^{8} - 4 x ^{2} + 8 c}{8 x ^{4}}; x \neq = 0

4 x + 6 = 2 x - 8

6 \cdot (2 - x) + 4 = 1 - (x - 3)

\frac{2 x}{5} = - 4

3 x = nπ + \frac{π}{2}

\frac{7}{6} - \frac{1}{2} = \frac{13}{12}

What is the answer to solvefor y,x^4y=(x^8)/8-(x^2)/2+c ?
The answer to solvefor y,x^4y=(x^8)/8-(x^2)/2+c is y=(x^8-4x^2+8c)/(8x^4);\quad x\ne 0