Calculus Examples

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

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

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

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

Step 1.3

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

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

Step 1.4

Since $3 x$ is constant with respect to $y$ , the derivative of $3 x$ with respect to $y$ is $0$ .

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

Step 1.5

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

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

Step 1.6

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

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

Step 1.7

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

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

Step 1.8

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

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

Step 1.9

Multiply $- 1$ by $1$ .

$\frac{\partial M}{\partial y} = 2 x (1 - e^{- x^{2}})$

Step 1.10

Simplify.

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Step 1.10.1

Apply the distributive property.

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

Step 1.10.2

Combine terms.

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Step 1.10.2.1

Multiply $2$ by $1$ .

$\frac{\partial M}{\partial y} = 2 x + 2 x (- e^{- x^{2}})$

Step 1.10.2.2

Multiply $- 1$ by $2$ .

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

Step 1.10.3

Reorder terms.

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

Step 1.10.4

Reorder factors in $- 2 e^{- x^{2}} x + 2 x$ .

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

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

Differentiate.

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.2.3

Since $3 y^{2}$ is constant with respect to $x$ , the derivative of $3 y^{2}$ with respect to $x$ is $0$ .

$\frac{\partial N}{\partial x} = 2 x + 0 + \frac{d}{d x} [e^{- x^{2}}]$

Step 2.3

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

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Step 2.3.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = - x^{2}$ .

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Step 2.3.1.1

To apply the Chain Rule, set $u$ as $- x^{2}$ .

$\frac{\partial N}{\partial x} = 2 x + 0 + \frac{d}{d u} [e^{u}] \frac{d}{d x} [- x^{2}]$

Step 2.3.1.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

$\frac{\partial N}{\partial x} = 2 x + 0 + e^{u} \frac{d}{d x} [- x^{2}]$

Step 2.3.1.3

Replace all occurrences of $u$ with $- x^{2}$ .

$\frac{\partial N}{\partial x} = 2 x + 0 + e^{- x^{2}} \frac{d}{d x} [- x^{2}]$

Step 2.3.2

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

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

Step 2.3.3

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

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

Step 2.3.4

Multiply $2$ by $- 1$ .

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

Step 2.4

Simplify.

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Step 2.4.1

Add $2 x$ and $0$ .

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

Step 2.4.2

Reorder terms.

$\frac{\partial N}{\partial x} = - 2 e^{- x^{2}} x + 2 x$

Step 2.4.3

Reorder factors in $- 2 e^{- x^{2}} x + 2 x$ .

$\frac{\partial N}{\partial x} = - 2 x e^{- x^{2}} + 2 x$

Step 3

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

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Step 3.1

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

$- 2 x e^{- x^{2}} + 2 x = - 2 x e^{- x^{2}} + 2 x$

Step 3.2

Since the two sides have been shown to be equivalent, the equation is an identity.

$- 2 x e^{- x^{2}} + 2 x = - 2 x e^{- x^{2}} + 2 x$ is an identity.

Step 4

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

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

Step 5

Integrate $N (x, y) = x^{2} + 3 y^{2} + e^{- x^{2}}$ to find $f (x, y)$ .

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Step 5.1

Split the single integral into multiple integrals.

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

Step 5.2

Apply the constant rule.

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

Step 5.3

Since $3$ is constant with respect to $y$ , move $3$ out of the integral.

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

Step 5.4

By the Power Rule, the integral of $y^{2}$ with respect to $y$ is $\frac{1}{3} y^{3}$ .

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

Step 5.5

Apply the constant rule.

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

Step 5.6

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

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

Step 5.7

Simplify.

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

Step 6

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

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

Step 7

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

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

Step 8

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

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Step 8.1

Differentiate $f$ with respect to $x$ .

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

Step 8.2

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

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

Step 8.3

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

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Step 8.3.1

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

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

Step 8.3.2

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

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

Step 8.3.3

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

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

Step 8.4

Since $y^{3}$ is constant with respect to $x$ , the derivative of $y^{3}$ with respect to $x$ is $0$ .

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

Step 8.5

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

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Step 8.5.1

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

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

Step 8.5.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = - x^{2}$ .

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Step 8.5.2.1

To apply the Chain Rule, set $u$ as $- x^{2}$ .

$2 y x + 0 + y (\frac{d}{d u} [e^{u}] \frac{d}{d x} [- x^{2}]) + \frac{d}{d x} [g (x)] = 2 x (3 x + y - y e^{- x^{2}})$

Step 8.5.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

$2 y x + 0 + y (e^{u} \frac{d}{d x} [- x^{2}]) + \frac{d}{d x} [g (x)] = 2 x (3 x + y - y e^{- x^{2}})$

Step 8.5.2.3

Replace all occurrences of $u$ with $- x^{2}$ .

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

Step 8.5.3

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

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

Step 8.5.4

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

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

Step 8.5.5

Multiply $2$ by $- 1$ .

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

Step 8.6

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

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

Step 8.7

Simplify.

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Step 8.7.1

Add $2 y x$ and $0$ .

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

Step 8.7.2

Reorder terms.

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

Step 8.7.3

Reorder factors in $\frac{d g}{d x} + 2 x y - 2 e^{- x^{2}} x y$ .

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

Step 9

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

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Step 9.1

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

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Step 9.1.1

Simplify $2 x (3 x + y - y e^{- x^{2}})$ .

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Step 9.1.1.1

Rewrite.

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

Step 9.1.1.2

Simplify by adding zeros.

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

Step 9.1.1.3

Apply the distributive property.

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

Step 9.1.1.4

Simplify.

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Step 9.1.1.4.1

Rewrite using the commutative property of multiplication.

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

Step 9.1.1.4.2

Multiply $- 1$ by $2$ .

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

Step 9.1.1.5

Simplify each term.

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Step 9.1.1.5.1

Multiply $x$ by $x$ by adding the exponents.

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Step 9.1.1.5.1.1

Move $x$ .

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

Step 9.1.1.5.1.2

Multiply $x$ by $x$ .

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

Step 9.1.1.5.2

Multiply $2$ by $3$ .

$\frac{d g}{d x} + 2 x y - 2 x y e^{- x^{2}} = 6 x^{2} + 2 x y - 2 x (y e^{- x^{2}})$

Step 9.1.1.6

Reorder factors in $6 x^{2} + 2 x y - 2 x (y e^{- x^{2}})$ .

$\frac{d g}{d x} + 2 x y - 2 x y e^{- x^{2}} = 6 x^{2} + 2 x y - 2 x y e^{- x^{2}}$

Step 9.1.2

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

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Step 9.1.2.1

Subtract $2 x y$ from both sides of the equation.

$\frac{d g}{d x} - 2 x y e^{- x^{2}} = 6 x^{2} + 2 x y - 2 x y e^{- x^{2}} - 2 x y$

Step 9.1.2.2

Add $2 x y e^{- x^{2}}$ to both sides of the equation.

$\frac{d g}{d x} = 6 x^{2} + 2 x y - 2 x y e^{- x^{2}} - 2 x y + 2 x y e^{- x^{2}}$

Step 9.1.2.3

Combine the opposite terms in $6 x^{2} + 2 x y - 2 x y e^{- x^{2}} - 2 x y + 2 x y e^{- x^{2}}$ .

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Step 9.1.2.3.1

Subtract $2 x y$ from $2 x y$ .

$\frac{d g}{d x} = 6 x^{2} - 2 x y e^{- x^{2}} + 0 + 2 x y e^{- x^{2}}$

Step 9.1.2.3.2

Add $6 x^{2} - 2 x y e^{- x^{2}}$ and $0$ .

$\frac{d g}{d x} = 6 x^{2} - 2 x y e^{- x^{2}} + 2 x y e^{- x^{2}}$

Step 9.1.2.3.3

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

$\frac{d g}{d x} = 6 x^{2} + 0$

Step 9.1.2.3.4

Add $6 x^{2}$ and $0$ .

$\frac{d g}{d x} = 6 x^{2}$

Step 10

Find the antiderivative of $6 x^{2}$ to find $g (x)$ .

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Step 10.1

Integrate both sides of $\frac{d g}{d x} = 6 x^{2}$ .

$\int \frac{d g}{d x} d x = \int 6 x^{2} d x$

Step 10.2

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

$g (x) = \int 6 x^{2} d x$

Step 10.3

Since $6$ is constant with respect to $x$ , move $6$ out of the integral.

$g (x) = 6 \int x^{2} d x$

Step 10.4

By the Power Rule, the integral of $x^{2}$ with respect to $x$ is $\frac{1}{3} x^{3}$ .

$g (x) = 6 (\frac{1}{3} x^{3} + C)$

Step 10.5

Simplify the answer.

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Step 10.5.1

Rewrite $6 (\frac{1}{3} x^{3} + C)$ as $6 (\frac{1}{3}) x^{3} + C$ .

$g (x) = 6 (\frac{1}{3}) x^{3} + C$

Step 10.5.2

Simplify.

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Step 10.5.2.1

Combine $6$ and $\frac{1}{3}$ .

$g (x) = \frac{6}{3} x^{3} + C$

Step 10.5.2.2

Cancel the common factor of $6$ and $3$ .

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Step 10.5.2.2.1

Factor $3$ out of $6$ .

$g (x) = \frac{3 \cdot 2}{3} x^{3} + C$

Step 10.5.2.2.2

Cancel the common factors.

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Step 10.5.2.2.2.1

Factor $3$ out of $3$ .

$g (x) = \frac{3 \cdot 2}{3 (1)} x^{3} + C$

Step 10.5.2.2.2.2

Cancel the common factor.

$g (x) = \frac{3 \cdot 2}{3 \cdot 1} x^{3} + C$

Step 10.5.2.2.2.3

Rewrite the expression.

$g (x) = \frac{2}{1} x^{3} + C$

Step 10.5.2.2.2.4

Divide $2$ by $1$ .

$g (x) = 2 x^{3} + C$

Step 11

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

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

Step 12

Reorder factors in $f (x, y) = x^{2} y + y^{3} + e^{- x^{2}} y + 2 x^{3} + C$ .

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