Calculus Examples

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

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

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

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

Step 1.3

Evaluate $\frac{d}{d y} [3 e^{x} y]$ .

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Step 1.3.1

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

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

Step 1.3.2

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

$\frac{\partial M}{\partial y} = 3 e^{x} \cdot 1 + \frac{d}{d y} [x]$

Step 1.3.3

Multiply $3$ by $1$ .

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

Step 1.4

Differentiate using the Constant Rule.

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Step 1.4.1

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

$\frac{\partial M}{\partial y} = 3 e^{x} + 0$

Step 1.4.2

Add $3 e^{x}$ and $0$ .

$\frac{\partial M}{\partial y} = 3 e^{x}$

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

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 $3 e^{x}$ for $\frac{\partial M}{\partial y}$ and $e^{x}$ for $\frac{\partial N}{\partial x}$ .

$3 e^{x} = e^{x}$

Step 3.2

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

$3 e^{x} = e^{x}$ is not an identity.

Step 4

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

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Step 4.1

Substitute $3 e^{x}$ for $\frac{\partial M}{\partial y}$ .

$\frac{3 e^{x} - \frac{\partial N}{\partial x}}{N}$

Step 4.2

Substitute $e^{x}$ for $\frac{\partial N}{\partial x}$ .

$\frac{3 e^{x} - e^{x}}{N}$

Step 4.3

Substitute $e^{x}$ for $N$ .

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Step 4.3.1

Substitute $e^{x}$ for $N$ .

$\frac{3 e^{x} - e^{x}}{e^{x}}$

Step 4.3.2

Subtract $e^{x}$ from $3 e^{x}$ .

$\frac{2 e^{x}}{e^{x}}$

Step 4.3.3

Cancel the common factor of $e^{x}$ .

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Step 4.3.3.1

Cancel the common factor.

$\frac{2 e^{x}}{e^{x}}$

Step 4.3.3.2

Divide $2$ by $1$ .

$2$

Step 4.4

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

$μ (x, y) = e^{\int 2 d x}$

Step 5

Evaluate the integral $e^{\int 2 d x}$ .

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Step 5.1

Apply the constant rule.

$μ (x, y) = e^{2 x + C}$

Step 5.2

Simplify.

$μ (x, y) = e^{2 x} + C$

Step 6

Multiply both sides of $(3 e^{x} y + x) d x + e^{x} d y = 0$ by the integration factor $e^{2 x}$ .

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Step 6.1

Multiply $3 e^{x} y + x$ by $e^{2 x}$ .

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

Step 6.2

Apply the distributive property.

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

Step 6.3

Multiply $e^{x}$ by $e^{2 x}$ by adding the exponents.

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Step 6.3.1

Move $e^{2 x}$ .

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

Step 6.3.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 6.3.3

Add $2 x$ and $x$ .

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

Step 6.4

Reorder factors in $3 e^{3 x} y + x e^{2 x}$ .

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

Step 6.5

Multiply $e^{x}$ by $e^{2 x}$ .

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

Step 6.6

Multiply $e^{x}$ by $e^{2 x}$ by adding the exponents.

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Step 6.6.1

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 6.6.2

Add $x$ and $2 x$ .

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

Step 7

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

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

Step 8

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

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Step 8.1

Apply the constant rule.

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

Step 9

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

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

Step 10

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

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

Step 11

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

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Step 11.1

Differentiate $f$ with respect to $x$ .

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

Step 11.2

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

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

Step 11.3

Evaluate $\frac{d}{d x} [e^{3 x} y]$ .

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Step 11.3.1

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

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

Step 11.3.2

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 $g (x) = 3 x$ .

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Step 11.3.2.1

To apply the Chain Rule, set $u$ as $3 x$ .

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

Step 11.3.2.2

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

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

Step 11.3.2.3

Replace all occurrences of $u$ with $3 x$ .

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

Step 11.3.3

Since $3$ is constant with respect to $x$ , the derivative of $3 x$ with respect to $x$ is $3 \frac{d}{d x} [x]$ .

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

Step 11.3.4

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

$y (e^{3 x} (3 \cdot 1)) + \frac{d}{d x} [g (x)] = 3 y e^{3 x} + x e^{2 x}$

Step 11.3.5

Multiply $3$ by $1$ .

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

Step 11.3.6

Move $3$ to the left of $e^{3 x}$ .

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

Step 11.3.7

Move $3$ to the left of $y$ .

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

Step 11.4

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

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

Step 11.5

Simplify.

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Step 11.5.1

Reorder terms.

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

Step 11.5.2

Reorder factors in $\frac{d g}{d x} + 3 e^{3 x} y$ .

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

Step 12

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

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Step 12.1

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

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Step 12.1.1

Subtract $3 y e^{3 x}$ from both sides of the equation.

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

Step 12.1.2

Combine the opposite terms in $3 y e^{3 x} + x e^{2 x} - 3 y e^{3 x}$ .

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Step 12.1.2.1

Subtract $3 y e^{3 x}$ from $3 y e^{3 x}$ .

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

Step 12.1.2.2

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

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

Step 13

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

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Step 13.1

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

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

Step 13.2

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

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

Step 13.3

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = x$ and $d v = e^{2 x}$ .

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

Step 13.4

Simplify.

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Step 13.4.1

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

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

Step 13.4.2

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

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

Step 13.5

Since $\frac{1}{2}$ is constant with respect to $x$ , move $\frac{1}{2}$ out of the integral.

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

Step 13.6

Remove parentheses.

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

Step 13.7

Let $u = 2 x$ . Then $d u = 2 d x$ , so $\frac{1}{2} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 13.7.1

Let $u = 2 x$ . Find $\frac{d u}{d x}$ .

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Step 13.7.1.1

Differentiate $2 x$ .

$\frac{d}{d x} [2 x]$

Step 13.7.1.2

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

$2 \frac{d}{d x} [x]$

Step 13.7.1.3

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

$2 \cdot 1$

Step 13.7.1.4

Multiply $2$ by $1$ .

$2$

Step 13.7.2

Rewrite the problem using $u$ and $d u$ .

$g (x) = \frac{x e^{2 x}}{2} - \frac{1}{2} \int e^{u} \frac{1}{2} d u$

Step 13.8

Combine $e^{u}$ and $\frac{1}{2}$ .

$g (x) = \frac{x e^{2 x}}{2} - \frac{1}{2} \int \frac{e^{u}}{2} d u$

Step 13.9

Since $\frac{1}{2}$ is constant with respect to $u$ , move $\frac{1}{2}$ out of the integral.

$g (x) = \frac{x e^{2 x}}{2} - \frac{1}{2} (\frac{1}{2} \int e^{u} d u)$

Step 13.10

Simplify.

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Step 13.10.1

Multiply $\frac{1}{2}$ by $\frac{1}{2}$ .

$g (x) = \frac{x e^{2 x}}{2} - \frac{1}{2 \cdot 2} \int e^{u} d u$

Step 13.10.2

Multiply $2$ by $2$ .

$g (x) = \frac{x e^{2 x}}{2} - \frac{1}{4} \int e^{u} d u$

Step 13.11

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

$g (x) = \frac{x e^{2 x}}{2} - \frac{1}{4} (e^{u} + C)$

Step 13.12

Rewrite $\frac{x e^{2 x}}{2} - \frac{1}{4} (e^{u} + C)$ as $\frac{1}{2} x e^{2 x} - \frac{1}{4} e^{u} + C$ .

$g (x) = \frac{1}{2} x e^{2 x} - \frac{1}{4} e^{u} + C$

Step 13.13

Replace all occurrences of $u$ with $2 x$ .

$g (x) = \frac{1}{2} x e^{2 x} - \frac{1}{4} e^{2 x} + C$

Step 14

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

$f (x, y) = e^{3 x} y + \frac{1}{2} x e^{2 x} - \frac{1}{4} e^{2 x} + C$

Step 15

Simplify $f (x, y) = e^{3 x} y + \frac{1}{2} x e^{2 x} - \frac{1}{4} e^{2 x} + C$ .

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Step 15.1

Simplify each term.

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Step 15.1.1

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

$f (x, y) = e^{3 x} y + \frac{x}{2} e^{2 x} - \frac{1}{4} e^{2 x} + C$

Step 15.1.2

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

$f (x, y) = e^{3 x} y + \frac{x e^{2 x}}{2} - \frac{1}{4} e^{2 x} + C$

Step 15.1.3

Combine $e^{2 x}$ and $\frac{1}{4}$ .

$f (x, y) = e^{3 x} y + \frac{x e^{2 x}}{2} - \frac{e^{2 x}}{4} + C$

Step 15.2

Reorder factors in $f (x, y) = e^{3 x} y + \frac{x e^{2 x}}{2} - \frac{e^{2 x}}{4} + C$ .

$f (x, y) = y e^{3 x} + \frac{x e^{2 x}}{2} - \frac{e^{2 x}}{4} + C$