Calculus Examples

$(y + 2 x) d y + d x = 0$

Step 1

Rewrite the differential equation to fit the Exact differential equation technique.

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Step 1.1

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

$d x + (y + 2 x) d y = 0$

Step 2

Find $\frac{\partial M}{\partial y}$ where $M (x, y) = 1$ .

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Step 2.1

Differentiate $M$ with respect to $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [1]$

Step 2.2

Since $1$ is constant with respect to $y$ , the derivative of $1$ with respect to $y$ is $0$ .

$\frac{\partial M}{\partial y} = 0$

Step 3

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

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Step 3.1

Differentiate $N$ with respect to $x$ .

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

Step 3.2

Differentiate.

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Step 3.2.1

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

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

Step 3.2.2

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

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

Step 3.3

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

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Step 3.3.1

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

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

Step 3.3.2

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

$\frac{\partial N}{\partial x} = 0 + 2 \cdot 1$

Step 3.3.3

Multiply $2$ by $1$ .

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

Step 3.4

Add $0$ and $2$ .

$\frac{\partial N}{\partial x} = 2$

Step 4

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

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Step 4.1

Substitute $0$ for $\frac{\partial M}{\partial y}$ and $2$ for $\frac{\partial N}{\partial x}$ .

$0 = 2$

Step 4.2

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

$0 = 2$ is not an identity.

Step 5

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 5.1

Substitute $2$ for $\frac{\partial N}{\partial x}$ .

$\frac{2 - \frac{\partial M}{\partial y}}{M}$

Step 5.2

Substitute $0$ for $\frac{\partial M}{\partial y}$ .

$\frac{2 + 0}{M}$

Step 5.3

Substitute $1$ for $M$ .

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Step 5.3.1

Substitute $1$ for $M$ .

$\frac{2 + 0}{1}$

Step 5.3.2

Divide $2 + 0$ by $1$ .

$2 + 0$

Step 5.3.3

Substitute $1$ for $M$ .

$2$

Step 5.4

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

$μ (x, y) = e^{\int 2 d y}$

Step 6

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

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Step 6.1

Apply the constant rule.

$μ (x, y) = e^{2 y + C}$

Step 6.2

Simplify.

$μ (x, y) = e^{2 y} + C$

Step 7

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

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Step 7.1

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

$d x + (y + 2 x) d y = 0$

Step 7.2

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

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

Step 7.3

Apply the distributive property.

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

Step 8

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

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

Step 9

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

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Step 9.1

Apply the constant rule.

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

Step 10

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

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

Step 11

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

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

Step 12

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

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Step 12.1

Differentiate $f$ with respect to $y$ .

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

Step 12.2

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

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

Step 12.3

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

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Step 12.3.1

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

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

Step 12.3.2

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

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Step 12.3.2.1

To apply the Chain Rule, set $u$ as $2 y$ .

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

Step 12.3.2.2

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

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

Step 12.3.2.3

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

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

Step 12.3.3

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

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

Step 12.3.4

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

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

Step 12.3.5

Multiply $2$ by $1$ .

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

Step 12.3.6

Move $2$ to the left of $e^{2 y}$ .

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

Step 12.3.7

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

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

Step 12.4

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

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

Step 12.5

Simplify.

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Step 12.5.1

Reorder terms.

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

Step 12.5.2

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

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

Step 13

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

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Step 13.1

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

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Step 13.1.1

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

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

Step 13.1.2

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

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Step 13.1.2.1

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

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

Step 13.1.2.2

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

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

Step 14

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

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Step 14.1

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

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

Step 14.2

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

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

Step 14.3

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

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

Step 14.4

Simplify.

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Step 14.4.1

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

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

Step 14.4.2

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

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

Step 14.5

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

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

Step 14.6

Remove parentheses.

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

Step 14.7

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

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Step 14.7.1

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

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Step 14.7.1.1

Differentiate $2 y$ .

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

Step 14.7.1.2

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

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

Step 14.7.1.3

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

$2 \cdot 1$

Step 14.7.1.4

Multiply $2$ by $1$ .

$2$

Step 14.7.2

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

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

Step 14.8

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

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

Step 14.9

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

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

Step 14.10

Simplify.

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Step 14.10.1

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

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

Step 14.10.2

Multiply $2$ by $2$ .

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

Step 14.11

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

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

Step 14.12

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

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

Step 14.13

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

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

Step 15

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

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

Step 16

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

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Step 16.1

Simplify each term.

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Step 16.1.1

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

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

Step 16.1.2

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

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

Step 16.1.3

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

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

Step 16.2

Subtract $\frac{e^{2 y}}{4}$ from $e^{2 y} x$ .

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Step 16.2.1

Reorder $e^{2 y}$ and $x$ .

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

Step 16.2.2

To write $x e^{2 y}$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

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

Step 16.2.3

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

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

Step 16.2.4

Combine the numerators over the common denominator.

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

Step 16.3

Simplify the numerator.

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Step 16.3.1

Move $4$ to the left of $x e^{2 y}$ .

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

Step 16.3.2

Factor $e^{2 y}$ out of $4 x e^{2 y} - e^{2 y}$ .

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Step 16.3.2.1

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

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

Step 16.3.2.2

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

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

Step 16.3.2.3

Factor $e^{2 y}$ out of $e^{2 y} (4 x) + e^{2 y} \cdot - 1$ .

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

Step 16.4

To write $\frac{y e^{2 y}}{2}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 16.5

Write each expression with a common denominator of $4$ , by multiplying each by an appropriate factor of $1$ .

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Step 16.5.1

Multiply $\frac{y e^{2 y}}{2}$ by $\frac{2}{2}$ .

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

Step 16.5.2

Multiply $2$ by $2$ .

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

Step 16.6

Combine the numerators over the common denominator.

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

Step 16.7

Simplify the numerator.

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Step 16.7.1

Factor $e^{2 y}$ out of $y e^{2 y} \cdot 2 + e^{2 y} (4 x - 1)$ .

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Step 16.7.1.1

Factor $e^{2 y}$ out of $y e^{2 y} \cdot 2$ .

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

Step 16.7.1.2

Factor $e^{2 y}$ out of $e^{2 y} (y \cdot 2) + e^{2 y} (4 x - 1)$ .

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

Step 16.7.2

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

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