Calculus Examples

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

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

Differentiate using the Product Rule which states that $\frac{d}{d y} [f (y) g (y)]$ is $f (y) \frac{d}{d y} [g (y)] + g (y) \frac{d}{d y} [f (y)]$ where $f (y) = y$ and $g (y) = 2 x - y$ .

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

Step 1.3

Differentiate.

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Step 1.3.1

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

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

Step 1.3.2

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

$\frac{\partial M}{\partial y} = y (0 + \frac{d}{d y} [- y]) + (2 x - y) \frac{d}{d y} [y]$

Step 1.3.3

Add $0$ and $\frac{d}{d y} [- y]$ .

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

Step 1.3.4

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

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

Step 1.3.5

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

$\frac{\partial M}{\partial y} = y (- 1 \cdot 1) + (2 x - y) \frac{d}{d y} [y]$

Step 1.3.6

Simplify the expression.

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Step 1.3.6.1

Multiply $- 1$ by $1$ .

$\frac{\partial M}{\partial y} = y \cdot - 1 + (2 x - y) \frac{d}{d y} [y]$

Step 1.3.6.2

Move $- 1$ to the left of $y$ .

$\frac{\partial M}{\partial y} = - 1 \cdot y + (2 x - y) \frac{d}{d y} [y]$

Step 1.3.6.3

Rewrite $- 1 y$ as $- y$ .

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

Step 1.3.7

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

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

Step 1.3.8

Simplify by adding terms.

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Step 1.3.8.1

Multiply $2 x - y$ by $1$ .

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

Step 1.3.8.2

Subtract $y$ from $- y$ .

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

Step 2

Find $\frac{\partial N}{\partial x}$ where $N (x, y) = - (x^{2} + y^{2} + 1)$ .

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Step 2.1

Differentiate $N$ with respect to $x$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [- (x^{2} + y^{2} + 1)]$

Step 2.2

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

$\frac{\partial N}{\partial x} = - \frac{d}{d x} [x^{2} + y^{2} + 1]$

Step 2.3

By the Sum Rule, the derivative of $x^{2} + y^{2} + 1$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [y^{2}] + \frac{d}{d x} [1]$ .

$\frac{\partial N}{\partial x} = - (\frac{d}{d x} [x^{2}] + \frac{d}{d x} [y^{2}] + \frac{d}{d x} [1])$

Step 2.4

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

$\frac{\partial N}{\partial x} = - (2 x + \frac{d}{d x} [y^{2}] + \frac{d}{d x} [1])$

Step 2.5

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

$\frac{\partial N}{\partial x} = - (2 x + 0 + \frac{d}{d x} [1])$

Step 2.6

Add $2 x$ and $0$ .

$\frac{\partial N}{\partial x} = - (2 x + \frac{d}{d x} [1])$

Step 2.7

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$\frac{\partial N}{\partial x} = - (2 x + 0)$

Step 2.8

Simplify the expression.

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Step 2.8.1

Add $2 x$ and $0$ .

$\frac{\partial N}{\partial x} = - (2 x)$

Step 2.8.2

Multiply $2$ by $- 1$ .

$\frac{\partial N}{\partial x} = - 2 x$

Step 3

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

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Step 3.1

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

$2 x - 2 y = - 2 x$

Step 3.2

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

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

Step 4

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 4.1

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

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

Step 4.2

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

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

Step 4.3

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

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Step 4.3.1

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

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

Step 4.3.2

Simplify the numerator.

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Step 4.3.2.1

Apply the distributive property.

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

Step 4.3.2.2

Multiply $2$ by $- 1$ .

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

Step 4.3.2.3

Multiply $- 2$ by $- 1$ .

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

Step 4.3.2.4

Subtract $2 x$ from $- 2 x$ .

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

Step 4.3.2.5

Factor $2$ out of $- 4 x + 2 y$ .

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Step 4.3.2.5.1

Factor $2$ out of $- 4 x$ .

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

Step 4.3.2.5.2

Factor $2$ out of $2 y$ .

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

Step 4.3.2.5.3

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

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

Step 4.3.3

Cancel the common factor of $- 2 x + y$ and $2 x - y$ .

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Step 4.3.3.1

Factor $- 1$ out of $- 2 x$ .

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

Step 4.3.3.2

Factor $- 1$ out of $y$ .

$\frac{2 (- (2 x) - 1 (- y))}{y (2 x - y)}$

Step 4.3.3.3

Factor $- 1$ out of $- (2 x) - 1 (- y)$ .

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

Step 4.3.3.4

Rewrite $- (2 x - y)$ as $- 1 (2 x - y)$ .

$\frac{2 (- 1 (2 x - y))}{y (2 x - y)}$

Step 4.3.3.5

Cancel the common factor.

$\frac{2 (- 1 (2 x - y))}{y (2 x - y)}$

Step 4.3.3.6

Rewrite the expression.

$\frac{2 \cdot (- 1)}{y}$

Step 4.3.4

Multiply $2$ by $- 1$ .

$\frac{- 2}{y}$

Step 4.3.5

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

$- \frac{2}{y}$

Step 4.4

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

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

Step 5

Evaluate the integral $e^{\int - \frac{2}{y} d y}$ .

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Step 5.1

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

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

Step 5.2

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

$μ (x, y) = e^{- (2 \int \frac{1}{y} d y)}$

Step 5.3

Multiply $2$ by $- 1$ .

$μ (x, y) = e^{- 2 \int \frac{1}{y} d y}$

Step 5.4

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

$μ (x, y) = e^{- 2 (\ln (| y |) + C)}$

Step 5.5

Simplify.

$μ (x, y) = e^{- 2 \ln (| y |)} + C$

Step 5.6

Simplify each term.

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Step 5.6.1

Simplify $- 2 \ln (| y |)$ by moving $- 2$ inside the logarithm.

$μ (x, y) = e^{\ln ({| y |}^{- 2})} + C$

Step 5.6.2

Exponentiation and log are inverse functions.

$μ (x, y) = {| y |}^{- 2} + C$

Step 5.6.3

Remove the absolute value in ${| y |}^{- 2}$ because exponentiations with even powers are always positive.

$μ (x, y) = y^{- 2} + C$

Step 5.6.4

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

$μ (x, y) = \frac{1}{y^{2}} + C$

Step 6

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

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Step 6.1

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

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

Step 6.2

Cancel the common factor of $y$ .

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Step 6.2.1

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

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

Step 6.2.2

Cancel the common factor.

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

Step 6.2.3

Rewrite the expression.

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

Step 6.3

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

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

Step 6.4

Multiply $- (x^{2} + y^{2} + 1)$ by $\frac{1}{y^{2}}$ .

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

Step 6.5

Apply the distributive property.

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

Step 6.6

Multiply $- 1$ by $1$ .

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

Step 6.7

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

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

Step 6.8

Factor $- 1$ out of $- x^{2}$ .

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

Step 6.9

Factor $- 1$ out of $- y^{2}$ .

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

Step 6.10

Factor $- 1$ out of $- (x^{2}) - (y^{2})$ .

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

Step 6.11

Rewrite $- 1$ as $- 1 (1)$ .

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

Step 6.12

Factor $- 1$ out of $- (x^{2} + y^{2}) - 1 (1)$ .

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

Step 6.13

Rewrite $- (x^{2} + y^{2} + 1)$ as $- 1 (x^{2} + y^{2} + 1)$ .

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

Step 6.14

Move the negative in front of the fraction.

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

Step 7

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

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

Step 8

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

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Step 8.1

Split the fraction into multiple fractions.

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

Step 8.2

Split the single integral into multiple integrals.

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

Step 8.3

Cancel the common factor of $y$ .

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Step 8.3.1

Cancel the common factor.

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

Step 8.3.2

Divide $- 1$ by $1$ .

$f (x, y) = \int \frac{2 x}{y} d x + \int - 1 d x$

Step 8.4

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

$f (x, y) = \frac{2}{y} \int x d x + \int - 1 d x$

Step 8.5

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

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

Step 8.6

Apply the constant rule.

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

Step 8.7

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

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

Step 8.8

Simplify.

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

Step 8.9

Simplify.

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Step 8.9.1

Move $2$ to the left of $y$ .

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

Step 8.9.2

Multiply $2$ by $y$ .

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

Step 8.9.3

Cancel the common factor of $2$ .

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Step 8.9.3.1

Cancel the common factor.

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

Step 8.9.3.2

Rewrite the expression.

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

Step 9

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

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

Step 10

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

$\frac{\partial f}{\partial y} = - \frac{x^{2} + y^{2} + 1}{y^{2}}$

Step 11

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

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Step 11.1

Differentiate $f$ with respect to $y$ .

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

Step 11.2

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

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

Step 11.3

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

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Step 11.3.1

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

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

Step 11.3.2

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

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

Step 11.3.3

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

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

Step 11.4

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

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

Step 11.5

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

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

Step 11.6

Simplify.

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Step 11.6.1

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

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

Step 11.6.2

Combine terms.

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Step 11.6.2.1

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

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

Step 11.6.2.2

Add $- \frac{x^{2}}{y^{2}}$ and $0$ .

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

Step 11.6.3

Reorder terms.

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

Step 12

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

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Step 12.1

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

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Step 12.1.1

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

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Step 12.1.1.1

Add $\frac{x^{2} + y^{2} + 1}{y^{2}}$ to both sides of the equation.

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

Step 12.1.1.2

Combine the numerators over the common denominator.

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

Step 12.1.1.3

Add $- x^{2}$ and $x^{2}$ .

$\frac{d g}{d y} + \frac{0 + y^{2} + 1}{y^{2}} = 0$

Step 12.1.1.4

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

$\frac{d g}{d y} + \frac{y^{2} + 1}{y^{2}} = 0$

Step 12.1.1.5

To write $\frac{d g}{d y}$ as a fraction with a common denominator, multiply by $\frac{y^{2}}{y^{2}}$ .

$\frac{\frac{d g}{d y} y^{2}}{y^{2}} + \frac{y^{2} + 1}{y^{2}} = 0$

Step 12.1.1.6

Combine the numerators over the common denominator.

$\frac{\frac{d g}{d y} y^{2} + y^{2} + 1}{y^{2}} = 0$

Step 12.1.2

Set the numerator equal to zero.

$\frac{d g}{d y} y^{2} + y^{2} + 1 = 0$

Step 12.1.3

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

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Step 12.1.3.1

Move all terms not containing $\frac{d g}{d y}$ to the right side of the equation.

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Step 12.1.3.1.1

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

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

Step 12.1.3.1.2

Subtract $1$ from both sides of the equation.

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

Step 12.1.3.2

Divide each term in $\frac{d g}{d y} y^{2} = - y^{2} - 1$ by $y^{2}$ and simplify.

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Step 12.1.3.2.1

Divide each term in $\frac{d g}{d y} y^{2} = - y^{2} - 1$ by $y^{2}$ .

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

Step 12.1.3.2.2

Simplify the left side.

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Step 12.1.3.2.2.1

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

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Step 12.1.3.2.2.1.1

Cancel the common factor.

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

Step 12.1.3.2.2.1.2

Divide $\frac{d g}{d y}$ by $1$ .

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

Step 12.1.3.2.3

Simplify the right side.

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Step 12.1.3.2.3.1

Simplify each term.

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Step 12.1.3.2.3.1.1

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

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Step 12.1.3.2.3.1.1.1

Cancel the common factor.

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

Step 12.1.3.2.3.1.1.2

Divide $- 1$ by $1$ .

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

Step 12.1.3.2.3.1.2

Move the negative in front of the fraction.

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

Step 13

Find the antiderivative of $- 1 - \frac{1}{y^{2}}$ to find $g (y)$ .

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Step 13.1

Integrate both sides of $\frac{d g}{d y} = - 1 - \frac{1}{y^{2}}$ .

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

Step 13.2

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

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

Step 13.3

Split the single integral into multiple integrals.

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

Step 13.4

Apply the constant rule.

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

Step 13.5

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

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

Step 13.6

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

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

Step 13.7

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

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Step 13.7.1

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

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

Step 13.7.2

Multiply $2$ by $- 1$ .

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

Step 13.8

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

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

Step 13.9

Simplify.

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

Step 13.10

Simplify.

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Step 13.10.1

Multiply $- 1$ by $- 1$ .

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

Step 13.10.2

Multiply $\frac{1}{y}$ by $1$ .

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

Step 14

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

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