Calculus Examples

$(3 x^{2} - 2 y^{2}) d y - 2 x y d x = 0$

Step 1

Rewrite the differential equation to fit the Exact differential equation technique.

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Step 1.1

Rewrite.

$- 2 x y d x + (3 x^{2} - 2 y^{2}) d y = 0$

Step 2

Find $\frac{\partial M}{\partial y}$ where $M (x, y) = - 2 x y$ .

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Step 2.1

Differentiate $M$ with respect to $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [- 2 x y]$

Step 2.2

Since $- 2 x$ is constant with respect to $y$ , the derivative of $- 2 x y$ with respect to $y$ is $- 2 x \frac{d}{d y} [y]$ .

$\frac{\partial M}{\partial y} = - 2 x \frac{d}{d y} [y]$

Step 2.3

Differentiate using the Power Rule which states that $\frac{d}{d y} [y^{n}]$ is $n y^{n - 1}$ where $n = 1$ .

$\frac{\partial M}{\partial y} = - 2 x \cdot 1$

Step 2.4

Multiply $- 2$ by $1$ .

$\frac{\partial M}{\partial y} = - 2 x$

Step 3

Find $\frac{\partial N}{\partial x}$ where $N (x, y) = 3 x^{2} - 2 y^{2}$ .

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Step 3.1

Differentiate $N$ with respect to $x$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [3 x^{2} - 2 y^{2}]$

Step 3.2

By the Sum Rule, the derivative of $3 x^{2} - 2 y^{2}$ with respect to $x$ is $\frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [- 2 y^{2}]$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [- 2 y^{2}]$

Step 3.3

Evaluate $\frac{d}{d x} [3 x^{2}]$ .

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Step 3.3.1

Since $3$ is constant with respect to $x$ , the derivative of $3 x^{2}$ with respect to $x$ is $3 \frac{d}{d x} [x^{2}]$ .

$\frac{\partial N}{\partial x} = 3 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 2 y^{2}]$

Step 3.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$\frac{\partial N}{\partial x} = 3 (2 x) + \frac{d}{d x} [- 2 y^{2}]$

Step 3.3.3

Multiply $2$ by $3$ .

$\frac{\partial N}{\partial x} = 6 x + \frac{d}{d x} [- 2 y^{2}]$

Step 3.4

Differentiate using the Constant Rule.

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Step 3.4.1

Since $- 2 y^{2}$ is constant with respect to $x$ , the derivative of $- 2 y^{2}$ with respect to $x$ is $0$ .

$\frac{\partial N}{\partial x} = 6 x + 0$

Step 3.4.2

Add $6 x$ and $0$ .

$\frac{\partial N}{\partial x} = 6 x$

Step 4

Check that $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ .

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Step 4.1

Substitute $- 2 x$ for $\frac{\partial M}{\partial y}$ and $6 x$ for $\frac{\partial N}{\partial x}$ .

$- 2 x = 6 x$

Step 4.2

Since the left side does not equal the right side, the equation is not an identity.

$- 2 x = 6 x$ is not an identity.

Step 5

Find the integration factor $μ (x, y) = e^{\int \frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{M} d y}$ .

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Step 5.1

Substitute $6 x$ for $\frac{\partial N}{\partial x}$ .

$\frac{6 x - \frac{\partial M}{\partial y}}{M}$

Step 5.2

Substitute $- 2 x$ for $\frac{\partial M}{\partial y}$ .

$\frac{6 x - (- 2 x)}{M}$

Step 5.3

Substitute $- 2 x y$ for $M$ .

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Step 5.3.1

Substitute $- 2 x y$ for $M$ .

$\frac{6 x - (- 2 x)}{- 2 x y}$

Step 5.3.2

Simplify the numerator.

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Step 5.3.2.1

Factor $x$ out of $6 x - (- 2 x)$ .

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Step 5.3.2.1.1

Factor $x$ out of $6 x$ .

$\frac{x \cdot 6 - (- 2 x)}{- 2 x y}$

Step 5.3.2.1.2

Factor $x$ out of $- (- 2 x)$ .

$\frac{x \cdot 6 + x (- (- 2))}{- 2 x y}$

Step 5.3.2.1.3

Factor $x$ out of $x \cdot 6 + x (- (- 2))$ .

$\frac{x (6 - (- 2))}{- 2 x y}$

Step 5.3.2.2

Multiply $- 1$ by $- 2$ .

$\frac{x (6 + 2)}{- 2 x y}$

Step 5.3.2.3

Add $6$ and $2$ .

$\frac{x \cdot 8}{- 2 x y}$

Step 5.3.3

Cancel the common factor of $x$ .

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Step 5.3.3.1

Cancel the common factor.

$\frac{x \cdot 8}{- 2 x y}$

Step 5.3.3.2

Rewrite the expression.

$\frac{8}{- 2 y}$

Step 5.3.4

Cancel the common factor of $8$ and $- 2$ .

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Step 5.3.4.1

Factor $2$ out of $8$ .

$\frac{2 (4)}{- 2 y}$

Step 5.3.4.2

Cancel the common factors.

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Step 5.3.4.2.1

Factor $2$ out of $- 2 y$ .

$\frac{2 (4)}{2 (- y)}$

Step 5.3.4.2.2

Cancel the common factor.

$\frac{2 \cdot 4}{2 (- y)}$

Step 5.3.4.2.3

Rewrite the expression.

$\frac{4}{- y}$

Step 5.3.5

Substitute $- 2 x y$ for $M$ .

$- \frac{4}{y}$

Step 5.4

Find the integration factor $μ (x, y) = e^{\int \frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{M} d y}$ .

$μ (x, y) = e^{\int - \frac{4}{y} d y}$

Step 6

Evaluate the integral $e^{\int - \frac{4}{y} d y}$ .

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Step 6.1

Since $- 1$ is constant with respect to $y$ , move $- 1$ out of the integral.

$μ (x, y) = e^{- \int \frac{4}{y} d y}$

Step 6.2

Since $4$ is constant with respect to $y$ , move $4$ out of the integral.

$μ (x, y) = e^{- (4 \int \frac{1}{y} d y)}$

Step 6.3

Multiply $4$ by $- 1$ .

$μ (x, y) = e^{- 4 \int \frac{1}{y} d y}$

Step 6.4

The integral of $\frac{1}{y}$ with respect to $y$ is $\ln (| y |)$ .

$μ (x, y) = e^{- 4 (\ln (| y |) + C)}$

Step 6.5

Simplify.

$μ (x, y) = e^{- 4 \ln (| y |)} + C$

Step 6.6

Simplify each term.

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Step 6.6.1

Simplify $- 4 \ln (| y |)$ by moving $- 4$ inside the logarithm.

$μ (x, y) = e^{\ln ({| y |}^{- 4})} + C$

Step 6.6.2

Exponentiation and log are inverse functions.

$μ (x, y) = {| y |}^{- 4} + C$

Step 6.6.3

Remove the absolute value in ${| y |}^{- 4}$ because exponentiations with even powers are always positive.

$μ (x, y) = y^{- 4} + C$

Step 6.6.4

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$μ (x, y) = \frac{1}{y^{4}} + C$

Step 7

Multiply both sides of $- 2 x y d x + (3 x^{2} - 2 y^{2}) d y = 0$ by the integration factor $\frac{1}{y^{4}}$ .

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Step 7.1

Multiply $- 2 x y$ by $\frac{1}{y^{4}}$ .

$- 2 x y \frac{1}{y^{4}} d x + (3 x^{2} - 2 y^{2}) d y = 0$

Step 7.2

Cancel the common factor of $y$ .

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Step 7.2.1

Factor $y$ out of $- 2 x y$ .

$y (- 2 x) \frac{1}{y^{4}} d x + (3 x^{2} - 2 y^{2}) d y = 0$

Step 7.2.2

Factor $y$ out of $y^{4}$ .

$y (- 2 x) \frac{1}{y \cdot y^{3}} d x + (3 x^{2} - 2 y^{2}) d y = 0$

Step 7.2.3

Cancel the common factor.

$y (- 2 x) \frac{1}{y \cdot y^{3}} d x + (3 x^{2} - 2 y^{2}) d y = 0$

Step 7.2.4

Rewrite the expression.

$- 2 x \frac{1}{y^{3}} d x + (3 x^{2} - 2 y^{2}) d y = 0$

Step 7.3

Combine $- 2$ and $\frac{1}{y^{3}}$ .

$x \frac{- 2}{y^{3}} d x + (3 x^{2} - 2 y^{2}) d y = 0$

Step 7.4

Combine $x$ and $\frac{- 2}{y^{3}}$ .

$\frac{x \cdot - 2}{y^{3}} d x + (3 x^{2} - 2 y^{2}) d y = 0$

Step 7.5

Move $- 2$ to the left of $x$ .

$\frac{- 2 \cdot x}{y^{3}} d x + (3 x^{2} - 2 y^{2}) d y = 0$

Step 7.6

Move the negative in front of the fraction.

$- \frac{2 x}{y^{3}} d x + (3 x^{2} - 2 y^{2}) d y = 0$

Step 7.7

Multiply $3 x^{2} - 2 y^{2}$ by $\frac{1}{y^{4}}$ .

$- \frac{2 x}{y^{3}} d x + (3 x^{2} - 2 y^{2}) \frac{1}{y^{4}} d y = 0$

Step 7.8

Multiply $3 x^{2} - 2 y^{2}$ by $\frac{1}{y^{4}}$ .

$- \frac{2 x}{y^{3}} d x + \frac{3 x^{2} - 2 y^{2}}{y^{4}} d y = 0$

Step 8

Set $f (x, y)$ equal to the integral of $M (x, y)$ .

$f (x, y) = \int - \frac{2 x}{y^{3}} d x$

Step 9

Integrate $M (x, y) = - \frac{2 x}{y^{3}}$ to find $f (x, y)$ .

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Step 9.1

Since $- 1$ is constant with respect to $x$ , move $- 1$ out of the integral.

$f (x, y) = - \int \frac{2 x}{y^{3}} d x$

Step 9.2

Since $\frac{2}{y^{3}}$ is constant with respect to $x$ , move $\frac{2}{y^{3}}$ out of the integral.

$f (x, y) = - (\frac{2}{y^{3}} \int x d x)$

Step 9.3

Remove parentheses.

$f (x, y) = - \frac{2}{y^{3}} \int x d x$

Step 9.4

By the Power Rule, the integral of $x$ with respect to $x$ is $\frac{1}{2} x^{2}$ .

$f (x, y) = - \frac{2}{y^{3}} (\frac{1}{2} x^{2} + C)$

Step 9.5

Simplify the answer.

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Step 9.5.1

Rewrite $- \frac{2}{y^{3}} (\frac{1}{2} x^{2} + C)$ as $- \frac{2}{y^{3}} \cdot \frac{1}{2} x^{2} + C$ .

$f (x, y) = - \frac{2}{y^{3}} \cdot \frac{1}{2} x^{2} + C$

Step 9.5.2

Simplify.

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Step 9.5.2.1

Multiply $\frac{1}{2}$ by $\frac{2}{y^{3}}$ .

$f (x, y) = - \frac{2}{2 y^{3}} x^{2} + C$

Step 9.5.2.2

Cancel the common factor of $2$ .

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Step 9.5.2.2.1

Cancel the common factor.

$f (x, y) = - \frac{2}{2 y^{3}} x^{2} + C$

Step 9.5.2.2.2

Rewrite the expression.

$f (x, y) = - \frac{1}{y^{3}} x^{2} + C$

Step 9.5.2.3

Combine $x^{2}$ and $\frac{1}{y^{3}}$ .

$f (x, y) = - \frac{x^{2}}{y^{3}} + C$

Step 10

Since the integral of $g (y)$ will contain an integration constant, we can replace $C$ with $g (y)$ .

$f (x, y) = - \frac{x^{2}}{y^{3}} + g (y)$

Step 11

Set $\frac{\partial f}{\partial y} = N (x, y)$ .

$\frac{\partial f}{\partial y} = \frac{3 x^{2} - 2 y^{2}}{y^{4}}$

Step 12

Find $\frac{\partial f}{\partial y}$ .

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Step 12.1

Differentiate $f$ with respect to $y$ .

$\frac{d}{d y} [- \frac{x^{2}}{y^{3}} + g (y)] = \frac{3 x^{2} - 2 y^{2}}{y^{4}}$

Step 12.2

By the Sum Rule, the derivative of $- \frac{x^{2}}{y^{3}} + g (y)$ with respect to $y$ is $\frac{d}{d y} [- \frac{x^{2}}{y^{3}}] + \frac{d}{d y} [g (y)]$ .

$\frac{d}{d y} [- \frac{x^{2}}{y^{3}}] + \frac{d}{d y} [g (y)] = \frac{3 x^{2} - 2 y^{2}}{y^{4}}$

Step 12.3

Evaluate $\frac{d}{d y} [- \frac{x^{2}}{y^{3}}]$ .

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Step 12.3.1

Since $- x^{2}$ is constant with respect to $y$ , the derivative of $- \frac{x^{2}}{y^{3}}$ with respect to $y$ is $- x^{2} \frac{d}{d y} [\frac{1}{y^{3}}]$ .

$- x^{2} \frac{d}{d y} [\frac{1}{y^{3}}] + \frac{d}{d y} [g (y)] = \frac{3 x^{2} - 2 y^{2}}{y^{4}}$

Step 12.3.2

Rewrite $\frac{1}{y^{3}}$ as ${(y^{3})}^{- 1}$ .

$- x^{2} \frac{d}{d y} [{(y^{3})}^{- 1}] + \frac{d}{d y} [g (y)] = \frac{3 x^{2} - 2 y^{2}}{y^{4}}$

Step 12.3.3

Differentiate using the chain rule, which states that $\frac{d}{d y} [f (g (y))]$ is $f' (g (y)) g' (y)$ where $f (y) = y^{- 1}$ and $g (y) = y^{3}$ .

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Step 12.3.3.1

To apply the Chain Rule, set $u$ as $y^{3}$ .

$- x^{2} (\frac{d}{d u} [u^{- 1}] \frac{d}{d y} [y^{3}]) + \frac{d}{d y} [g (y)] = \frac{3 x^{2} - 2 y^{2}}{y^{4}}$

Step 12.3.3.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = - 1$ .

$- x^{2} (- u^{- 2} \frac{d}{d y} [y^{3}]) + \frac{d}{d y} [g (y)] = \frac{3 x^{2} - 2 y^{2}}{y^{4}}$

Step 12.3.3.3

Replace all occurrences of $u$ with $y^{3}$ .

$- x^{2} (- {(y^{3})}^{- 2} \frac{d}{d y} [y^{3}]) + \frac{d}{d y} [g (y)] = \frac{3 x^{2} - 2 y^{2}}{y^{4}}$

Step 12.3.4

Differentiate using the Power Rule which states that $\frac{d}{d y} [y^{n}]$ is $n y^{n - 1}$ where $n = 3$ .

$- x^{2} (- {(y^{3})}^{- 2} (3 y^{2})) + \frac{d}{d y} [g (y)] = \frac{3 x^{2} - 2 y^{2}}{y^{4}}$

Step 12.3.5

Multiply the exponents in ${(y^{3})}^{- 2}$ .

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Step 12.3.5.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$- x^{2} (- y^{3 \cdot - 2} (3 y^{2})) + \frac{d}{d y} [g (y)] = \frac{3 x^{2} - 2 y^{2}}{y^{4}}$

Step 12.3.5.2

Multiply $3$ by $- 2$ .

$- x^{2} (- y^{- 6} (3 y^{2})) + \frac{d}{d y} [g (y)] = \frac{3 x^{2} - 2 y^{2}}{y^{4}}$

Step 12.3.6

Multiply $3$ by $- 1$ .

$- x^{2} (- 3 y^{- 6} y^{2}) + \frac{d}{d y} [g (y)] = \frac{3 x^{2} - 2 y^{2}}{y^{4}}$

Step 12.3.7

Multiply $y^{- 6}$ by $y^{2}$ by adding the exponents.

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Step 12.3.7.1

Move $y^{2}$ .

$- x^{2} (- 3 (y^{2} y^{- 6})) + \frac{d}{d y} [g (y)] = \frac{3 x^{2} - 2 y^{2}}{y^{4}}$

Step 12.3.7.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$- x^{2} (- 3 y^{2 - 6}) + \frac{d}{d y} [g (y)] = \frac{3 x^{2} - 2 y^{2}}{y^{4}}$

Step 12.3.7.3

Subtract $6$ from $2$ .

$- x^{2} (- 3 y^{- 4}) + \frac{d}{d y} [g (y)] = \frac{3 x^{2} - 2 y^{2}}{y^{4}}$

Step 12.3.8

Multiply $- 3$ by $- 1$ .

$3 x^{2} y^{- 4} + \frac{d}{d y} [g (y)] = \frac{3 x^{2} - 2 y^{2}}{y^{4}}$

Step 12.4

Differentiate using the function rule which states that the derivative of $g (y)$ is $\frac{d g}{d y}$ .

$3 x^{2} y^{- 4} + \frac{d g}{d y} = \frac{3 x^{2} - 2 y^{2}}{y^{4}}$

Step 12.5

Simplify.

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Step 12.5.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$3 x^{2} \frac{1}{y^{4}} + \frac{d g}{d y} = \frac{3 x^{2} - 2 y^{2}}{y^{4}}$

Step 12.5.2

Combine terms.

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Step 12.5.2.1

Combine $3$ and $\frac{1}{y^{4}}$ .

$x^{2} \frac{3}{y^{4}} + \frac{d g}{d y} = \frac{3 x^{2} - 2 y^{2}}{y^{4}}$

Step 12.5.2.2

Combine $x^{2}$ and $\frac{3}{y^{4}}$ .

$\frac{x^{2} \cdot 3}{y^{4}} + \frac{d g}{d y} = \frac{3 x^{2} - 2 y^{2}}{y^{4}}$

Step 12.5.2.3

Move $3$ to the left of $x^{2}$ .

$\frac{3 x^{2}}{y^{4}} + \frac{d g}{d y} = \frac{3 x^{2} - 2 y^{2}}{y^{4}}$

Step 12.5.3

Reorder terms.

$\frac{d g}{d y} + \frac{3 x^{2}}{y^{4}} = \frac{3 x^{2} - 2 y^{2}}{y^{4}}$

Step 13

Solve for $\frac{d g}{d y}$ .

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Step 13.1

Solve for $\frac{d g}{d y}$ .

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Step 13.1.1

Move all terms containing variables to the left side of the equation.

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Step 13.1.1.1

Subtract $\frac{3 x^{2} - 2 y^{2}}{y^{4}}$ from both sides of the equation.

$\frac{d g}{d y} + \frac{3 x^{2}}{y^{4}} - \frac{3 x^{2} - 2 y^{2}}{y^{4}} = 0$

Step 13.1.1.2

Combine the numerators over the common denominator.

$\frac{d g}{d y} + \frac{3 x^{2} - (3 x^{2} - 2 y^{2})}{y^{4}} = 0$

Step 13.1.1.3

Simplify each term.

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Step 13.1.1.3.1

Apply the distributive property.

$\frac{d g}{d y} + \frac{3 x^{2} - (3 x^{2}) - (- 2 y^{2})}{y^{4}} = 0$

Step 13.1.1.3.2

Multiply $3$ by $- 1$ .

$\frac{d g}{d y} + \frac{3 x^{2} - 3 x^{2} - (- 2 y^{2})}{y^{4}} = 0$

Step 13.1.1.3.3

Multiply $- 2$ by $- 1$ .

$\frac{d g}{d y} + \frac{3 x^{2} - 3 x^{2} + 2 y^{2}}{y^{4}} = 0$

Step 13.1.1.4

Subtract $3 x^{2}$ from $3 x^{2}$ .

$\frac{d g}{d y} + \frac{0 + 2 y^{2}}{y^{4}} = 0$

Step 13.1.1.5

Add $0$ and $2 y^{2}$ .

$\frac{d g}{d y} + \frac{2 y^{2}}{y^{4}} = 0$

Step 13.1.1.6

Cancel the common factor of $y^{2}$ and $y^{4}$ .

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Step 13.1.1.6.1

Factor $y^{2}$ out of $2 y^{2}$ .

$\frac{d g}{d y} + \frac{y^{2} \cdot 2}{y^{4}} = 0$

Step 13.1.1.6.2

Cancel the common factors.

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Step 13.1.1.6.2.1

Factor $y^{2}$ out of $y^{4}$ .

$\frac{d g}{d y} + \frac{y^{2} \cdot 2}{y^{2} y^{2}} = 0$

Step 13.1.1.6.2.2

Cancel the common factor.

$\frac{d g}{d y} + \frac{y^{2} \cdot 2}{y^{2} y^{2}} = 0$

Step 13.1.1.6.2.3

Rewrite the expression.

$\frac{d g}{d y} + \frac{2}{y^{2}} = 0$

Step 13.1.2

Subtract $\frac{2}{y^{2}}$ from both sides of the equation.

$\frac{d g}{d y} = - \frac{2}{y^{2}}$

Step 14

Find the antiderivative of $- \frac{2}{y^{2}}$ to find $g (y)$ .

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Step 14.1

Integrate both sides of $\frac{d g}{d y} = - \frac{2}{y^{2}}$ .

$\int \frac{d g}{d y} d y = \int - \frac{2}{y^{2}} d y$

Step 14.2

Evaluate $\int \frac{d g}{d y} d y$ .

$g (y) = \int - \frac{2}{y^{2}} d y$

Step 14.3

Since $- 1$ is constant with respect to $y$ , move $- 1$ out of the integral.

$g (y) = - \int \frac{2}{y^{2}} d y$

Step 14.4

Since $2$ is constant with respect to $y$ , move $2$ out of the integral.

$g (y) = - (2 \int \frac{1}{y^{2}} d y)$

Step 14.5

Multiply $2$ by $- 1$ .

$g (y) = - 2 \int \frac{1}{y^{2}} d y$

Step 14.6

Move $y^{2}$ out of the denominator by raising it to the $- 1$ power.

$g (y) = - 2 \int {(y^{2})}^{- 1} d y$

Step 14.7

Multiply the exponents in ${(y^{2})}^{- 1}$ .

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Step 14.7.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$g (y) = - 2 \int y^{2 \cdot - 1} d y$

Step 14.7.2

Multiply $2$ by $- 1$ .

$g (y) = - 2 \int y^{- 2} d y$

Step 14.8

By the Power Rule, the integral of $y^{- 2}$ with respect to $y$ is $- y^{- 1}$ .

$g (y) = - 2 (- y^{- 1} + C)$

Step 14.9

Simplify the answer.

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Step 14.9.1

Rewrite $- 2 (- y^{- 1} + C)$ as $- 2 (- \frac{1}{y}) + C$ .

$g (y) = - 2 (- \frac{1}{y}) + C$

Step 14.9.2

Simplify.

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Step 14.9.2.1

Multiply $- 1$ by $- 2$ .

$g (y) = 2 \frac{1}{y} + C$

Step 14.9.2.2

Combine $2$ and $\frac{1}{y}$ .

$g (y) = \frac{2}{y} + C$

Step 15

Substitute for $g (y)$ in $f (x, y) = - \frac{x^{2}}{y^{3}} + g (y)$ .

$f (x, y) = - \frac{x^{2}}{y^{3}} + \frac{2}{y} + C$