Calculus Examples

$y (4 x + y) d x - 2 (x^{2} - y) d y = 0$

Step 1

Find $\frac{\partial M}{\partial y}$ where $M (x, y) = y (4 x + y)$ .

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Step 1.1

Differentiate $M$ with respect to $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [y (4 x + y)]$

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 $g (y) = 4 x + y$ .

$\frac{\partial M}{\partial y} = y \frac{d}{d y} [4 x + y] + (4 x + y) \frac{d}{d y} [y]$

Step 1.3

Differentiate.

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Step 1.3.1

By the Sum Rule, the derivative of $4 x + y$ with respect to $y$ is $\frac{d}{d y} [4 x] + \frac{d}{d y} [y]$ .

$\frac{\partial M}{\partial y} = y (\frac{d}{d y} [4 x] + \frac{d}{d y} [y]) + (4 x + y) \frac{d}{d y} [y]$

Step 1.3.2

Since $4 x$ is constant with respect to $y$ , the derivative of $4 x$ with respect to $y$ is $0$ .

$\frac{\partial M}{\partial y} = y (0 + \frac{d}{d y} [y]) + (4 x + y) \frac{d}{d y} [y]$

Step 1.3.3

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

$\frac{\partial M}{\partial y} = y \frac{d}{d y} [y] + (4 x + y) \frac{d}{d y} [y]$

Step 1.3.4

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

$\frac{\partial M}{\partial y} = y \cdot 1 + (4 x + y) \frac{d}{d y} [y]$

Step 1.3.5

Multiply $y$ by $1$ .

$\frac{\partial M}{\partial y} = y + (4 x + y) \frac{d}{d y} [y]$

Step 1.3.6

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

$\frac{\partial M}{\partial y} = y + (4 x + y) \cdot 1$

Step 1.3.7

Simplify by adding terms.

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Step 1.3.7.1

Multiply $4 x + y$ by $1$ .

$\frac{\partial M}{\partial y} = y + 4 x + y$

Step 1.3.7.2

Add $y$ and $y$ .

$\frac{\partial M}{\partial y} = 4 x + 2 y$

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

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

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

Step 2.3

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

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

Step 2.4

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

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

Step 2.5

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

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

Step 2.6

Simplify the expression.

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Step 2.6.1

Add $2 x$ and $0$ .

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

Step 2.6.2

Multiply $2$ by $- 2$ .

$\frac{\partial N}{\partial x} = - 4 x$

Step 3

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

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Step 3.1

Substitute $4 x + 2 y$ for $\frac{\partial M}{\partial y}$ and $- 4 x$ for $\frac{\partial N}{\partial x}$ .

$4 x + 2 y = - 4 x$

Step 3.2

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

$4 x + 2 y = - 4 x$ 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 $- 4 x$ for $\frac{\partial N}{\partial x}$ .

$\frac{- 4 x - \frac{\partial M}{\partial y}}{M}$

Step 4.2

Substitute $4 x + 2 y$ for $\frac{\partial M}{\partial y}$ .

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

Step 4.3

Substitute $y (4 x + y)$ for $M$ .

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Step 4.3.1

Substitute $y (4 x + y)$ for $M$ .

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

Step 4.3.2

Simplify the numerator.

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Step 4.3.2.1

Apply the distributive property.

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

Step 4.3.2.2

Multiply $4$ by $- 1$ .

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

Step 4.3.2.3

Multiply $2$ by $- 1$ .

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

Step 4.3.2.4

Subtract $4 x$ from $- 4 x$ .

$\frac{- 8 x - 2 y}{y (4 x + y)}$

Step 4.3.2.5

Factor $2$ out of $- 8 x - 2 y$ .

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Step 4.3.2.5.1

Factor $2$ out of $- 8 x$ .

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

Step 4.3.2.5.2

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

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

Step 4.3.2.5.3

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

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

Step 4.3.3

Cancel the common factor of $- 4 x - y$ and $4 x + y$ .

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Step 4.3.3.1

Factor $- 1$ out of $- 4 x$ .

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

Step 4.3.3.2

Factor $- 1$ out of $- y$ .

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

Step 4.3.3.3

Factor $- 1$ out of $- (4 x) - (y)$ .

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

Step 4.3.3.4

Rewrite $- (4 x + y)$ as $- 1 (4 x + y)$ .

$\frac{2 (- 1 (4 x + y))}{y (4 x + y)}$

Step 4.3.3.5

Cancel the common factor.

$\frac{2 (- 1 (4 x + y))}{y (4 x + y)}$

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

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Step 6.1

Multiply $y (4 x + y)$ by $\frac{1}{y^{2}}$ .

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

Step 6.2

Cancel the common factor of $y$ .

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Step 6.2.1

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

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

Step 6.2.2

Cancel the common factor.

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

Step 6.2.3

Rewrite the expression.

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

Step 6.3

Multiply $4 x + y$ by $\frac{1}{y}$ .

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

Step 6.4

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

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

Step 6.5

Apply the distributive property.

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

Step 6.6

Multiply $- 1$ by $- 2$ .

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

Step 6.7

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

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

Step 6.8

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

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Step 6.8.1

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

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

Step 6.8.2

Factor $2$ out of $2 y$ .

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

Step 6.8.3

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

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

Step 6.9

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

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

Step 6.10

Factor $- 1$ out of $y$ .

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

Step 6.11

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

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

Step 6.12

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

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

Step 6.13

Move the negative in front of the fraction.

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

Step 7

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

$f (x, y) = \int \frac{4 x + y}{y} d x$

Step 8

Integrate $M (x, y) = \frac{4 x + y}{y}$ to find $f (x, y)$ .

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Step 8.1

Split the fraction into multiple fractions.

$f (x, y) = \int \frac{4 x}{y} + \frac{y}{y} d x$

Step 8.2

Split the single integral into multiple integrals.

$f (x, y) = \int \frac{4 x}{y} d x + \int \frac{y}{y} d x$

Step 8.3

Cancel the common factor of $y$ .

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Step 8.3.1

Cancel the common factor.

$f (x, y) = \int \frac{4 x}{y} d x + \int \frac{y}{y} d x$

Step 8.3.2

Rewrite the expression.

$f (x, y) = \int \frac{4 x}{y} d x + \int d x$

Step 8.4

Since $\frac{4}{y}$ is constant with respect to $x$ , move $\frac{4}{y}$ out of the integral.

$f (x, y) = \frac{4}{y} \int x d x + \int d x$

Step 8.5

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

$f (x, y) = \frac{4}{y} (\frac{1}{2} x^{2} + C) + \int d x$

Step 8.6

Apply the constant rule.

$f (x, y) = \frac{4}{y} (\frac{1}{2} x^{2} + C) + x + C$

Step 8.7

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

$f (x, y) = \frac{4}{y} (\frac{x^{2}}{2} + C) + x + C$

Step 8.8

Simplify.

$f (x, y) = \frac{4 x^{2}}{y \cdot 2} + x + C$

Step 8.9

Simplify.

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Step 8.9.1

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

$f (x, y) = \frac{4 x^{2}}{2 \cdot y} + x + C$

Step 8.9.2

Multiply $2$ by $y$ .

$f (x, y) = \frac{4 x^{2}}{2 y} + x + C$

Step 8.9.3

Cancel the common factor of $4$ and $2$ .

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Step 8.9.3.1

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

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

Step 8.9.3.2

Cancel the common factors.

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Step 8.9.3.2.1

Factor $2$ out of $2 y$ .

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

Step 8.9.3.2.2

Cancel the common factor.

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

Step 8.9.3.2.3

Rewrite the expression.

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

Step 9

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

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

Step 10

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

$\frac{\partial f}{\partial y} = - \frac{2 (x^{2} - y)}{y^{2}}$

Step 11

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

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Step 11.1

Differentiate $f$ with respect to $y$ .

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

Step 11.2

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

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

Step 11.3

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

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Step 11.3.1

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

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

Step 11.3.2

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

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

Step 11.3.3

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

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

Step 11.3.4

Multiply $- 1$ by $2$ .

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

Step 11.4

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

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

Step 11.5

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

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

Step 11.6

Simplify.

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Step 11.6.1

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

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

Step 11.6.2

Combine terms.

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Step 11.6.2.1

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

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

Step 11.6.2.2

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

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

Step 11.6.2.3

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

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

Step 11.6.2.4

Move the negative in front of the fraction.

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

Step 11.6.2.5

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

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

Step 11.6.3

Reorder terms.

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

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 $\frac{2 (x^{2} - y)}{y^{2}}$ to both sides of the equation.

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

Step 12.1.1.2

Combine the numerators over the common denominator.

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

Step 12.1.1.3

Simplify each term.

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Step 12.1.1.3.1

Apply the distributive property.

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

Step 12.1.1.3.2

Multiply $- 1$ by $2$ .

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

Step 12.1.1.4

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

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

Step 12.1.1.5

Subtract $2 y$ from $0$ .

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

Step 12.1.1.6

Simplify each term.

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Step 12.1.1.6.1

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

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Step 12.1.1.6.1.1

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

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

Step 12.1.1.6.1.2

Cancel the common factors.

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Step 12.1.1.6.1.2.1

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

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

Step 12.1.1.6.1.2.2

Cancel the common factor.

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

Step 12.1.1.6.1.2.3

Rewrite the expression.

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

Step 12.1.1.6.2

Move the negative in front of the fraction.

$\frac{d g}{d y} - \frac{2}{y} = 0$

Step 12.1.2

Add $\frac{2}{y}$ to both sides of the equation.

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

Step 13

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

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Step 13.1

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

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

Step 13.2

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

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

Step 13.3

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

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

Step 13.4

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

$g (y) = 2 (\ln (| y |) + C)$

Step 13.5

Simplify.

$g (y) = 2 \ln (| y |) + C$

Step 14

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

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

Step 15

Simplify each term.

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Step 15.1

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

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

Step 15.2

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

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