Calculus Examples

$x (2 y + 1) d y + 2 (y^{2} + y + 2 x^{2}) d x = 0$

Step 1

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

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Step 1.1

Rewrite.

$2 (y^{2} + y + 2 x^{2}) d x + x (2 y + 1) d y = 0$

Step 2

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

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Step 2.1

Differentiate $M$ with respect to $y$ .

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

Step 2.2

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

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

Step 2.3

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

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

Step 2.4

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

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

Step 2.5

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

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

Step 2.6

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

$\frac{\partial M}{\partial y} = 2 (2 y + 1 + 0)$

Step 2.7

Add $2 y + 1$ and $0$ .

$\frac{\partial M}{\partial y} = 2 (2 y + 1)$

Step 2.8

Simplify.

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Step 2.8.1

Apply the distributive property.

$\frac{\partial M}{\partial y} = 2 (2 y) + 2 \cdot 1$

Step 2.8.2

Combine terms.

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Step 2.8.2.1

Multiply $2$ by $2$ .

$\frac{\partial M}{\partial y} = 4 y + 2 \cdot 1$

Step 2.8.2.2

Multiply $2$ by $1$ .

$\frac{\partial M}{\partial y} = 4 y + 2$

Step 3

Find $\frac{\partial N}{\partial x}$ where $N (x, y) = x (2 y + 1)$ .

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Step 3.1

Differentiate $N$ with respect to $x$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [x (2 y + 1)]$

Step 3.2

Since $2 y + 1$ is constant with respect to $x$ , the derivative of $x (2 y + 1)$ with respect to $x$ is $(2 y + 1) \frac{d}{d x} [x]$ .

$\frac{\partial N}{\partial x} = (2 y + 1) \frac{d}{d x} [x]$

Step 3.3

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

$\frac{\partial N}{\partial x} = (2 y + 1) \cdot 1$

Step 3.4

Multiply $2 y + 1$ by $1$ .

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

Step 4

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

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Step 4.1

Substitute $4 y + 2$ for $\frac{\partial M}{\partial y}$ and $2 y + 1$ for $\frac{\partial N}{\partial x}$ .

$4 y + 2 = 2 y + 1$

Step 4.2

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

$4 y + 2 = 2 y + 1$ is not an identity.

Step 5

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 5.1

Substitute $4 y + 2$ for $\frac{\partial M}{\partial y}$ .

$\frac{4 y + 2 - \frac{\partial N}{\partial x}}{N}$

Step 5.2

Substitute $2 y + 1$ for $\frac{\partial N}{\partial x}$ .

$\frac{4 y + 2 - (2 y + 1)}{N}$

Step 5.3

Substitute $x (2 y + 1)$ for $N$ .

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Step 5.3.1

Substitute $x (2 y + 1)$ for $N$ .

$\frac{4 y + 2 - (2 y + 1)}{x (2 y + 1)}$

Step 5.3.2

Simplify the numerator.

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Step 5.3.2.1

Apply the distributive property.

$\frac{4 y + 2 - (2 y) - 1 \cdot 1}{x (2 y + 1)}$

Step 5.3.2.2

Multiply $2$ by $- 1$ .

$\frac{4 y + 2 - 2 y - 1 \cdot 1}{x (2 y + 1)}$

Step 5.3.2.3

Multiply $- 1$ by $1$ .

$\frac{4 y + 2 - 2 y - 1}{x (2 y + 1)}$

Step 5.3.2.4

Subtract $2 y$ from $4 y$ .

$\frac{2 y + 2 - 1}{x (2 y + 1)}$

Step 5.3.2.5

Subtract $1$ from $2$ .

$\frac{2 y + 1}{x (2 y + 1)}$

Step 5.3.3

Cancel the common factor of $2 y + 1$ .

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Step 5.3.3.1

Cancel the common factor.

$\frac{2 y + 1}{x (2 y + 1)}$

Step 5.3.3.2

Rewrite the expression.

$\frac{1}{x}$

Step 5.4

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

$μ (x, y) = e^{\int \frac{1}{x} d x}$

Step 6

Evaluate the integral $e^{\int \frac{1}{x} d x}$ .

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Step 6.1

The integral of $\frac{1}{x}$ with respect to $x$ is $\ln (| x |)$ .

$μ (x, y) = e^{\ln (| x |) + C}$

Step 6.2

Simplify the answer.

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Step 6.2.1

Simplify.

$μ (x, y) = e^{\ln (| x |)} + C$

Step 6.2.2

Exponentiation and log are inverse functions.

$μ (x, y) = | x | + C$

Step 7

Multiply both sides of $2 (y^{2} + y + 2 x^{2}) d x + x (2 y + 1) d y = 0$ by the integration factor $x$ .

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Step 7.1

Multiply $2 (y^{2} + y + 2 x^{2})$ by $x$ .

$2 (y^{2} + y + 2 x^{2}) x d x + x (2 y + 1) d y = 0$

Step 7.2

Apply the distributive property.

$(2 y^{2} + 2 y + 2 (2 x^{2})) x d x + x (2 y + 1) d y = 0$

Step 7.3

Multiply $2$ by $2$ .

$(2 y^{2} + 2 y + 4 x^{2}) x d x + x (2 y + 1) d y = 0$

Step 7.4

Apply the distributive property.

$(2 y^{2} x + 2 y x + 4 x^{2} x) d x + x (2 y + 1) d y = 0$

Step 7.5

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

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Step 7.5.1

Move $x$ .

$(2 y^{2} x + 2 y x + 4 (x \cdot x^{2})) d x + x (2 y + 1) d y = 0$

Step 7.5.2

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

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Step 7.5.2.1

Raise $x$ to the power of $1$ .

$(2 y^{2} x + 2 y x + 4 (x^{1} x^{2})) d x + x (2 y + 1) d y = 0$

Step 7.5.2.2

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

$(2 y^{2} x + 2 y x + 4 x^{1 + 2}) d x + x (2 y + 1) d y = 0$

Step 7.5.3

Add $1$ and $2$ .

$(2 y^{2} x + 2 y x + 4 x^{3}) d x + x (2 y + 1) d y = 0$

Step 7.6

Multiply $x (2 y + 1)$ by $x$ .

$(2 y^{2} x + 2 y x + 4 x^{3}) d x + x (2 y + 1) x d y = 0$

Step 7.7

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

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Step 7.7.1

Move $x$ .

$(2 y^{2} x + 2 y x + 4 x^{3}) d x + x \cdot x (2 y + 1) d y = 0$

Step 7.7.2

Multiply $x$ by $x$ .

$(2 y^{2} x + 2 y x + 4 x^{3}) d x + x^{2} (2 y + 1) d y = 0$

Step 7.8

Apply the distributive property.

$(2 y^{2} x + 2 y x + 4 x^{3}) d x + (x^{2} (2 y) + x^{2} \cdot 1) d y = 0$

Step 7.9

Rewrite using the commutative property of multiplication.

$(2 y^{2} x + 2 y x + 4 x^{3}) d x + (2 x^{2} y + x^{2} \cdot 1) d y = 0$

Step 7.10

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

$(2 y^{2} x + 2 y x + 4 x^{3}) d x + (2 x^{2} y + x^{2}) d y = 0$

Step 8

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

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

Step 9

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

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Step 9.1

Split the single integral into multiple integrals.

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

Step 9.2

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

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

Step 9.3

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

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

Step 9.4

Apply the constant rule.

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

Step 9.5

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

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

Step 9.6

Simplify.

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

Step 10

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

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

Step 11

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

$\frac{\partial f}{\partial x} = 2 y^{2} x + 2 y x + 4 x^{3}$

Step 12

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

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Step 12.1

Differentiate $f$ with respect to $x$ .

$\frac{d}{d x} [x^{2} y^{2} + x^{2} y + g (x)] = 2 y^{2} x + 2 y x + 4 x^{3}$

Step 12.2

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

$\frac{d}{d x} [x^{2} y^{2}] + \frac{d}{d x} [x^{2} y] + \frac{d}{d x} [g (x)] = 2 y^{2} x + 2 y x + 4 x^{3}$

Step 12.3

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

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Step 12.3.1

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

$y^{2} \frac{d}{d x} [x^{2}] + \frac{d}{d x} [x^{2} y] + \frac{d}{d x} [g (x)] = 2 y^{2} x + 2 y x + 4 x^{3}$

Step 12.3.2

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

$y^{2} (2 x) + \frac{d}{d x} [x^{2} y] + \frac{d}{d x} [g (x)] = 2 y^{2} x + 2 y x + 4 x^{3}$

Step 12.3.3

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

$2 y^{2} x + \frac{d}{d x} [x^{2} y] + \frac{d}{d x} [g (x)] = 2 y^{2} x + 2 y x + 4 x^{3}$

Step 12.4

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

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Step 12.4.1

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

$2 y^{2} x + y \frac{d}{d x} [x^{2}] + \frac{d}{d x} [g (x)] = 2 y^{2} x + 2 y x + 4 x^{3}$

Step 12.4.2

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

$2 y^{2} x + y (2 x) + \frac{d}{d x} [g (x)] = 2 y^{2} x + 2 y x + 4 x^{3}$

Step 12.4.3

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

$2 y^{2} x + 2 y x + \frac{d}{d x} [g (x)] = 2 y^{2} x + 2 y x + 4 x^{3}$

Step 12.5

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

$2 y^{2} x + 2 y x + \frac{d g}{d x} = 2 y^{2} x + 2 y x + 4 x^{3}$

Step 12.6

Reorder terms.

$\frac{d g}{d x} + 2 y^{2} x + 2 x y = 2 y^{2} x + 2 y x + 4 x^{3}$

Step 13

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

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Step 13.1

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

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Step 13.1.1

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

$\frac{d g}{d x} + 2 x y = 2 y^{2} x + 2 y x + 4 x^{3} - 2 y^{2} x$

Step 13.1.2

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

$\frac{d g}{d x} = 2 y^{2} x + 2 y x + 4 x^{3} - 2 y^{2} x - 2 x y$

Step 13.1.3

Combine the opposite terms in $2 y^{2} x + 2 y x + 4 x^{3} - 2 y^{2} x - 2 x y$ .

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Step 13.1.3.1

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

$\frac{d g}{d x} = 2 y x + 4 x^{3} + 0 - 2 x y$

Step 13.1.3.2

Add $2 y x + 4 x^{3}$ and $0$ .

$\frac{d g}{d x} = 2 y x + 4 x^{3} - 2 x y$

Step 13.1.3.3

Reorder the factors in the terms $2 y x$ and $- 2 x y$ .

$\frac{d g}{d x} = 2 x y + 4 x^{3} - 2 x y$

Step 13.1.3.4

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

$\frac{d g}{d x} = 4 x^{3} + 0$

Step 13.1.3.5

Add $4 x^{3}$ and $0$ .

$\frac{d g}{d x} = 4 x^{3}$

Step 14

Find the antiderivative of $4 x^{3}$ to find $g (x)$ .

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Step 14.1

Integrate both sides of $\frac{d g}{d x} = 4 x^{3}$ .

$\int \frac{d g}{d x} d x = \int 4 x^{3} d x$

Step 14.2

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

$g (x) = \int 4 x^{3} d x$

Step 14.3

Since $4$ is constant with respect to $x$ , move $4$ out of the integral.

$g (x) = 4 \int x^{3} d x$

Step 14.4

By the Power Rule, the integral of $x^{3}$ with respect to $x$ is $\frac{1}{4} x^{4}$ .

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

Step 14.5

Simplify the answer.

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Step 14.5.1

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

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

Step 14.5.2

Simplify.

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Step 14.5.2.1

Combine $4$ and $\frac{1}{4}$ .

$g (x) = \frac{4}{4} x^{4} + C$

Step 14.5.2.2

Cancel the common factor of $4$ .

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Step 14.5.2.2.1

Cancel the common factor.

$g (x) = \frac{4}{4} x^{4} + C$

Step 14.5.2.2.2

Rewrite the expression.

$g (x) = 1 x^{4} + C$

Step 14.5.2.3

Multiply $x^{4}$ by $1$ .

$g (x) = x^{4} + C$

Step 15

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

$f (x, y) = x^{2} y^{2} + x^{2} y + x^{4} + C$