Solve y = x - 2 x y - 2 x - y + 2 y + 1

Question

Solve the equation

Solve for x

Solve for y

x = \frac{1}{1 + 2 y}

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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

Not symmetry with respect to the origin

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Rewrite the equation

r = \frac{- cos ( θ ) + cos ^{2} ( θ ) + 4 sin ( 2 θ )}{2 sin ( 2 θ )} r = - \frac{cos ( θ ) + cos ^{2} ( θ ) + 4 sin ( 2 θ )}{2 sin ( 2 θ )}

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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{1 + 2 y}{2 x}

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Find the second derivative

Find the second derivative with respect to x

Find the second derivative with respect to y

\frac{d ^{2} y}{d x ^{2}} = \frac{1 + 2 y}{x ^{2}}

Calculate

y = x - 2 x y - 2 x - y + 2 y + 1

Simplify the expression

y = - x - 2 x y + y + 1

Take the derivative of both sides

\frac{d}{d x} (y) = \frac{d}{d x} (- x - 2 x y + y + 1)

Calculate the derivative

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\frac{d y}{d x} = \frac{d}{d x} (- x - 2 x y + y + 1)

Calculate the derivative

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\frac{d y}{d x} = - 1 - 2 y - 2 x \frac{d y}{d x} + \frac{d y}{d x}

Cancel equal terms on both sides of the expression

0 = - 1 - 2 y - 2 x \frac{d y}{d x}

Swap the sides of the equation

- 1 - 2 y - 2 x \frac{d y}{d x} = 0

Move the constant to the right side

- 2 x \frac{d y}{d x} = 0 + 1 + 2 y

Removing 0 doesn't change the value,so remove it from the expression

- 2 x \frac{d y}{d x} = 1 + 2 y

Divide both sides

\frac{- 2 x \frac{d y}{d x}}{- 2 x} = \frac{1 + 2 y}{- 2 x}

Divide the numbers

\frac{d y}{d x} = \frac{1 + 2 y}{- 2 x}

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

\frac{d y}{d x} = - \frac{1 + 2 y}{2 x}

Take the derivative of both sides

\frac{d}{d x} (\frac{d y}{d x}) = \frac{d}{d x} (- \frac{1 + 2 y}{2 x})

Calculate the derivative

\frac{d ^{2} y}{d x ^{2}} = \frac{d}{d x} (- \frac{1 + 2 y}{2 x})

Use differentiation rules

\frac{d ^{2} y}{d x ^{2}} = - \frac{\frac{d}{d x} ( 1 + 2 y ) \times 2 x - ( 1 + 2 y ) \times \frac{d}{d x} ( 2 x )}{( 2 x ) ^{2}}

Calculate the derivative

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\frac{d ^{2} y}{d x ^{2}} = - \frac{2 \frac{d y}{d x} \times 2 x - ( 1 + 2 y ) \times \frac{d}{d x} ( 2 x )}{( 2 x ) ^{2}}

Calculate the derivative

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\frac{d ^{2} y}{d x ^{2}} = - \frac{2 \frac{d y}{d x} \times 2 x - ( 1 + 2 y ) \times 2}{( 2 x ) ^{2}}

Calculate

\frac{d ^{2} y}{d x ^{2}} = - \frac{4 \frac{d y}{d x} \times x - ( 1 + 2 y ) \times 2}{( 2 x ) ^{2}}

Calculate

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

Calculate

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\frac{d ^{2} y}{d x ^{2}} = - \frac{4 x \frac{d y}{d x} - 2 - 4 y}{( 2 x ) ^{2}}

Calculate

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\frac{d ^{2} y}{d x ^{2}} = - \frac{4 x \frac{d y}{d x} - 2 - 4 y}{4 x ^{2}}

Calculate

\frac{d ^{2} y}{d x ^{2}} = - \frac{2 x \frac{d y}{d x} - 1 - 2 y}{2 x ^{2}}

Use equation \frac{d y}{d x} = - \frac{1 + 2 y}{2 x} to substitute

\frac{d ^{2} y}{d x ^{2}} = - \frac{2 x ( - \frac{1 + 2 y}{2 x} ) - 1 - 2 y}{2 x ^{2}}

Solution

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\frac{d ^{2} y}{d x ^{2}} = \frac{1 + 2 y}{x ^{2}}

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Conic

(x^{'} + \frac{2}{4})^{2} - (y^{'} + \frac{2}{4})^{2} = 1

Evaluate

y = x - 2 x y - 2 x - y + 2 y + 1

Move the expression to the left side

y - (x - 2 x y - 2 x - y + 2 y + 1) = 0

Calculate

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x + 2 x y - 1 = 0

The coefficients A,B and C of the general equation are A= 0,B= 2 and C= 0

A = 0 B = 2 C = 0

To find the angle of rotation θ,substitute the values of A,B and C into the formula cot (2 θ) = \frac{A - C}{B}

cot (2 θ) = \frac{0 - 0}{2}

Calculate

cot (2 θ) = 0

Using the unit circle,find the smallest positive angle for which the cotangent is 0

2 θ = \frac{π}{2}

Calculate

θ = \frac{π}{4}

To rotate the axes,use the equation of rotation and substitute \frac{π}{4} for θ

x = x^{'} cos (\frac{π}{4}) - y^{'} sin (\frac{π}{4}) y = x^{'} sin (\frac{π}{4}) + y^{'} cos (\frac{π}{4})

Calculate

x = x^{'} \times \frac{2}{2} - y^{'} sin (\frac{π}{4}) y = x^{'} sin (\frac{π}{4}) + y^{'} cos (\frac{π}{4})

Calculate

x = x^{'} \times \frac{2}{2} - y^{'} \times \frac{2}{2} y = x^{'} sin (\frac{π}{4}) + y^{'} cos (\frac{π}{4})

Calculate

x = x^{'} \times \frac{2}{2} - y^{'} \times \frac{2}{2} y = x^{'} \times \frac{2}{2} + y^{'} cos (\frac{π}{4})

Calculate

x = x^{'} \times \frac{2}{2} - y^{'} \times \frac{2}{2} y = x^{'} \times \frac{2}{2} + y^{'} \times \frac{2}{2}

Substitute x and y into the original equation x + 2 x y - 1 = 0

x^{'} \times \frac{2}{2} - y^{'} \times \frac{2}{2} + 2 (x^{'} \times \frac{2}{2} - y^{'} \times \frac{2}{2}) (x^{'} \times \frac{2}{2} + y^{'} \times \frac{2}{2}) - 1 = 0

Calculate

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\frac{2}{2} x^{'} - \frac{2}{2} y^{'} + (x^{'})^{2} - (y^{'})^{2} - 1 = 0

Move the constant to the right-hand side and change its sign

\frac{2}{2} x^{'} - \frac{2}{2} y^{'} + (x^{'})^{2} - (y^{'})^{2} = 0 - (- 1)

If a negative sign or a subtraction symbol appears outside parentheses, remove the parentheses and change the sign of every term within the parentheses

\frac{2}{2} x^{'} - \frac{2}{2} y^{'} + (x^{'})^{2} - (y^{'})^{2} = 0 + 1

Removing 0 doesn't change the value,so remove it from the expression

\frac{2}{2} x^{'} - \frac{2}{2} y^{'} + (x^{'})^{2} - (y^{'})^{2} = 1

Use the commutative property to reorder the terms

(x^{'})^{2} + \frac{2}{2} x^{'} - (y^{'})^{2} - \frac{2}{2} y^{'} = 1

To complete the square, the same value needs to be added to both sides

(x^{'})^{2} + \frac{2}{2} x^{'} + \frac{1}{8} - (y^{'})^{2} - \frac{2}{2} y^{'} = 1 + \frac{1}{8}

Use a^{2} + 2 ab + b^{2} = (a + b)^{2} to factor the expression

(x^{'} + \frac{2}{4})^{2} - (y^{'})^{2} - \frac{2}{2} y^{'} = 1 + \frac{1}{8}

Add the numbers

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(x^{'} + \frac{2}{4})^{2} - (y^{'})^{2} - \frac{2}{2} y^{'} = \frac{9}{8}

Factor out the negative sign from the expression

(x^{'} + \frac{2}{4})^{2} - (y^{'})^{2} - \frac{2}{2} y^{'} - \frac{1}{8} = \frac{9}{8} - \frac{1}{8}

Use a^{2} + 2 ab + b^{2} = (a + b)^{2} to factor the expression

(x^{'} + \frac{2}{4})^{2} - (y^{'} + \frac{2}{4})^{2} = \frac{9}{8} - \frac{1}{8}

Solution

More Steps

(x^{'} + \frac{2}{4})^{2} - (y^{'} + \frac{2}{4})^{2} = 1

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