Solve y^2 = -2(x^6 - x^3y - 32)

Question

Solve the equation

Solve for x

Solve for y

x = \frac{3 4 y + 4 - y ^{2} + 128}{2} x = \frac{3 4 y - 4 - y ^{2} + 128}{2}

Evaluate

y^{2} = - 2 (x^{6} - x^{3} y - 32)

Rewrite the expression

y^{2} = - 2 (x^{6} - y x^{3} - 32)

Swap the sides of the equation

- 2 (x^{6} - y x^{3} - 32) = y^{2}

Change the sign

2 (x^{6} - y x^{3} - 32) = - y^{2}

Divide both sides

\frac{2 ( x ^{6} - y x ^{3} - 32 )}{2} = \frac{- y ^{2}}{2}

Divide the numbers

x^{6} - y x^{3} - 32 = \frac{- y ^{2}}{2}

Use \frac{- a}{b} = \frac{a}{- b} = - \frac{a}{b} to rewrite the fraction

x^{6} - y x^{3} - 32 = - \frac{y ^{2}}{2}

Move the expression to the left side

x^{6} - y x^{3} - 32 - (- \frac{y ^{2}}{2}) = 0

Subtract the terms

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x^{6} - y x^{3} + \frac{- 64 + y ^{2}}{2} = 0

Multiply both sides of the equation by LCD

(x^{6} - y x^{3} + \frac{- 64 + y ^{2}}{2}) \times 2 = 0 \times 2

Simplify the equation

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2 x^{6} - 2 y x^{3} - 64 + y^{2} = 0 \times 2

Any expression multiplied by 0 equals 0

2 x^{6} - 2 y x^{3} - 64 + y^{2} = 0

Solve the equation using substitution t = x^{3}

2 t^{2} - 2 y t - 64 + y^{2} = 0

Substitute a= 2,b= - 2 y and c= - 64 + y^{2} into the quadratic formula t = \frac{- b \pm b ^{2} - 4 a c}{2 a}

t = \frac{2 y \pm ( - 2 y ) ^{2} - 4 \times 2 ( - 64 + y ^{2} )}{2 \times 2}

Simplify the expression

t = \frac{2 y \pm ( - 2 y ) ^{2} - 4 \times 2 ( - 64 + y ^{2} )}{4}

Simplify the expression

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t = \frac{2 y \pm - 4 y ^{2} + 512}{4}

Simplify the radical expression

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t = \frac{2 y \pm 2 - y ^{2} + 128}{4}

Separate the equation into 2 possible cases

t = \frac{2 y + 2 - y ^{2} + 128}{4} t = \frac{2 y - 2 - y ^{2} + 128}{4}

Simplify the expression

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t = \frac{y + - y ^{2} + 128}{2} t = \frac{2 y - 2 - y ^{2} + 128}{4}

Simplify the expression

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t = \frac{y + - y ^{2} + 128}{2} t = \frac{y - - y ^{2} + 128}{2}

Substitute back

x^{3} = \frac{y + - y ^{2} + 128}{2} x^{3} = \frac{y - - y ^{2} + 128}{2}

Solve the equation for x

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x = \frac{3 4 y + 4 - y ^{2} + 128}{2} x^{3} = \frac{y - - y ^{2} + 128}{2}

Solution

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x = \frac{3 4 y + 4 - y ^{2} + 128}{2} x = \frac{3 4 y - 4 - y ^{2} + 128}{2}

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Testing for symmetry

Testing for symmetry about the origin

Testing for symmetry about the x-axis

Testing for symmetry about the y-axis

Symmetry with respect to the origin

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Find the first derivative

Find the derivative with respect to x

Find the derivative with respect to y

\frac{d y}{d x} = \frac{- 6 x ^{5} + 3 y x ^{2}}{y - x ^{3}}

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Find the second derivative

\frac{d ^{2} y}{d x ^{2}} = \frac{- 33 y ^{2} x ^{4} + 48 x ^{7} y - 30 x ^{10} + 6 y ^{3} x}{y ^{3} - 3 y ^{2} x ^{3} + 3 y x ^{6} - x ^{9}}

Calculate

y^{2} = - 2 (x^{6} - x^{3} y - 32)

Take the derivative of both sides

\frac{d}{d x} (y^{2}) = \frac{d}{d x} (- 2 (x^{6} - x^{3} y - 32))

Calculate the derivative

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2 y \frac{d y}{d x} = \frac{d}{d x} (- 2 (x^{6} - x^{3} y - 32))

Calculate the derivative

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2 y \frac{d y}{d x} = - 12 x^{5} + 6 x^{2} y + 2 x^{3} \frac{d y}{d x}

Move the variable to the left side

2 y \frac{d y}{d x} - 2 x^{3} \frac{d y}{d x} = - 12 x^{5} + 6 x^{2} y

Collect like terms by calculating the sum or difference of their coefficients

(2 y - 2 x^{3}) \frac{d y}{d x} = - 12 x^{5} + 6 x^{2} y

Divide both sides

\frac{( 2 y - 2 x ^{3} ) \frac{d y}{d x}}{2 y - 2 x ^{3}} = \frac{- 12 x ^{5} + 6 x ^{2} y}{2 y - 2 x ^{3}}

Divide the numbers

\frac{d y}{d x} = \frac{- 12 x ^{5} + 6 x ^{2} y}{2 y - 2 x ^{3}}

Divide the numbers

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\frac{d y}{d x} = \frac{- 6 x ^{5} + 3 y x ^{2}}{y - x ^{3}}

Take the derivative of both sides

\frac{d}{d x} (\frac{d y}{d x}) = \frac{d}{d x} (\frac{- 6 x ^{5} + 3 y x ^{2}}{y - x ^{3}})

Calculate the derivative

\frac{d ^{2} y}{d x ^{2}} = \frac{d}{d x} (\frac{- 6 x ^{5} + 3 y x ^{2}}{y - x ^{3}})

Use differentiation rules

\frac{d ^{2} y}{d x ^{2}} = \frac{\frac{d}{d x} ( - 6 x ^{5} + 3 y x ^{2} ) \times ( y - x ^{3} ) - ( - 6 x ^{5} + 3 y x ^{2} ) \times \frac{d}{d x} ( y - x ^{3} )}{( y - x ^{3} ) ^{2}}

Calculate the derivative

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\frac{d ^{2} y}{d x ^{2}} = \frac{( - 30 x ^{4} + 6 x y + 3 x ^{2} \frac{d y}{d x} ) ( y - x ^{3} ) - ( - 6 x ^{5} + 3 y x ^{2} ) \times \frac{d}{d x} ( y - x ^{3} )}{( y - x ^{3} ) ^{2}}

Calculate the derivative

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\frac{d ^{2} y}{d x ^{2}} = \frac{( - 30 x ^{4} + 6 x y + 3 x ^{2} \frac{d y}{d x} ) ( y - x ^{3} ) - ( - 6 x ^{5} + 3 y x ^{2} ) ( \frac{d y}{d x} - 3 x ^{2} )}{( y - x ^{3} ) ^{2}}

Calculate

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\frac{d ^{2} y}{d x ^{2}} = \frac{- 36 x ^{4} y + 30 x ^{7} + 6 x y ^{2} + 3 x ^{2} y \frac{d y}{d x} - 3 x ^{5} \frac{d y}{d x} - ( - 6 x ^{5} + 3 y x ^{2} ) ( \frac{d y}{d x} - 3 x ^{2} )}{( y - x ^{3} ) ^{2}}

Calculate

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\frac{d ^{2} y}{d x ^{2}} = \frac{- 36 x ^{4} y + 30 x ^{7} + 6 x y ^{2} + 3 x ^{2} y \frac{d y}{d x} - 3 x ^{5} \frac{d y}{d x} - ( - 6 x ^{5} \frac{d y}{d x} + 3 y x ^{2} \frac{d y}{d x} + 18 x ^{7} - 9 y x ^{4} )}{( y - x ^{3} ) ^{2}}

Calculate

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\frac{d ^{2} y}{d x ^{2}} = \frac{- 27 x ^{4} y + 12 x ^{7} + 6 x y ^{2} + 3 x ^{5} \frac{d y}{d x}}{( y - x ^{3} ) ^{2}}

Use equation \frac{d y}{d x} = \frac{- 6 x ^{5} + 3 y x ^{2}}{y - x ^{3}} to substitute

\frac{d ^{2} y}{d x ^{2}} = \frac{- 27 x ^{4} y + 12 x ^{7} + 6 x y ^{2} + 3 x ^{5} \times \frac{- 6 x ^{5} + 3 y x ^{2}}{y - x ^{3}}}{( y - x ^{3} ) ^{2}}

Solution

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\frac{d ^{2} y}{d x ^{2}} = \frac{- 33 y ^{2} x ^{4} + 48 x ^{7} y - 30 x ^{10} + 6 y ^{3} x}{y ^{3} - 3 y ^{2} x ^{3} + 3 y x ^{6} - x ^{9}}

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