Calculus Examples

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

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

Differentiate.

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Step 1.2.1

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

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

Step 1.2.2

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

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

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

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

Step 1.3.2

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

$\frac{\partial M}{\partial y} = 0 - e^{x} \cdot 1$

Step 1.3.3

Multiply $- 1$ by $1$ .

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

Step 1.4

Subtract $e^{x}$ from $0$ .

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

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

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

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

Step 2.3

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

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

Step 3

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

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Step 3.1

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

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

Step 3.2

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

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

Step 4

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

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

Step 5

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

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Step 5.1

Apply the constant rule.

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

Step 6

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

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

Step 7

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

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

Step 8

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

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Step 8.1

Differentiate $f$ with respect to $x$ .

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

Step 8.2

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

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

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

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

Step 8.3.2

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

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

Step 8.4

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

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

Step 8.5

Simplify.

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Step 8.5.1

Reorder terms.

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

Step 8.5.2

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

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

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

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

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

Step 9.1.2

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

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Step 9.1.2.1

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

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

Step 9.1.2.2

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

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

Step 10

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

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Step 10.1

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

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

Step 10.2

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

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

Step 10.3

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

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

Step 10.4

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

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

Step 10.5

Simplify the answer.

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Step 10.5.1

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

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

Step 10.5.2

Combine $2$ and $\frac{1}{3}$ .

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

Step 11

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

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

Step 12

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

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Step 12.1

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

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

Step 12.2

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

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