Calculus Examples

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

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

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

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

Step 1.3

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

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Step 1.3.1

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

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

Step 1.3.2

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

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

Step 1.3.3

Multiply $e^{x}$ by $1$ .

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

Step 1.4

Differentiate.

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Step 1.4.1

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

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

Step 1.4.2

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

$\frac{\partial M}{\partial y} = e^{x} + 0 + 2 y$

Step 1.5

Simplify.

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Step 1.5.1

Add $e^{x}$ and $0$ .

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

Step 1.5.2

Reorder terms.

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

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

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

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

Step 2.3

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

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

Step 2.4

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

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Step 2.4.1

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

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

Step 2.4.2

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

$\frac{\partial N}{\partial x} = e^{x} + 2 y \cdot 1$

Step 2.4.3

Multiply $2$ by $1$ .

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

Step 2.5

Reorder terms.

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

Step 3

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

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Step 3.1

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

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

Step 3.2

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

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

Step 4

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

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

Step 5

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

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Step 5.1

Split the single integral into multiple integrals.

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

Step 5.2

Apply the constant rule.

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

Step 5.3

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

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

Step 5.4

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

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

Step 5.5

Simplify.

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

Step 5.6

Simplify.

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Step 5.6.1

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

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

Step 5.6.2

Cancel the common factor of $2$ .

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Step 5.6.2.1

Cancel the common factor.

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

Step 5.6.2.2

Rewrite the expression.

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

Step 5.6.3

Multiply $x$ by $1$ .

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

Step 6

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

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

Step 7

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

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

Step 8

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

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Step 8.1

Differentiate $f$ with respect to $x$ .

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

Step 8.2

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

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

Step 8.3

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

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Step 8.3.1

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

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

Step 8.3.2

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

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

Step 8.4

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

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Step 8.4.1

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

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

Step 8.4.2

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

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

Step 8.4.3

Multiply $y^{2}$ by $1$ .

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

Step 8.5

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

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

Step 8.6

Simplify.

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Step 8.6.1

Reorder terms.

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

Step 8.6.2

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

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

Step 9

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

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Step 9.1

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

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Step 9.1.1

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

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

Step 9.1.2

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

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

Step 9.1.3

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

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Step 9.1.3.1

Subtract $y^{2}$ from $y^{2}$ .

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

Step 9.1.3.2

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

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

Step 9.1.3.3

Subtract $y e^{x}$ from $y e^{x}$ .

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

Step 9.1.3.4

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

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

Step 10

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

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Step 10.1

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

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

Step 10.2

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

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

Step 10.3

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

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

Step 10.4

The integral of $e^{x}$ with respect to $x$ is $e^{x}$ .

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

Step 10.5

Simplify.

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

Step 11

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

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

Step 12

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

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