Calculus Examples

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

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

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

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

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

By the Sum Rule, the derivative of $2 y t + 1$ with respect to $x$ is $\frac{d}{d x} [2 y t] + \frac{d}{d x} [1]$ .

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

Step 2.3

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

$\frac{\partial N}{\partial x} = 0 + \frac{d}{d x} [1]$

Step 2.4

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

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

Step 2.5

Add $0$ and $0$ .

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

Step 3

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

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Step 3.1

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

$2 y = 0$

Step 3.2

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

$2 y = 0$ is not an identity.

Step 4

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 4.1

Substitute $0$ for $\frac{\partial N}{\partial x}$ .

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

Step 4.2

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

$\frac{0 - (2 y)}{M}$

Step 4.3

Substitute $y^{2}$ for $M$ .

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Step 4.3.1

Substitute $y^{2}$ for $M$ .

$\frac{0 - (2 y)}{y^{2}}$

Step 4.3.2

Simplify the numerator.

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Step 4.3.2.1

Multiply $- 1$ by $2$ .

$\frac{0 - 2 y}{y^{2}}$

Step 4.3.2.2

Subtract $2 y$ from $0$ .

$\frac{- 2 y}{y^{2}}$

Step 4.3.3

Cancel the common factor of $y$ and $y^{2}$ .

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Step 4.3.3.1

Factor $y$ out of $- 2 y$ .

$\frac{y \cdot - 2}{y^{2}}$

Step 4.3.3.2

Cancel the common factors.

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Step 4.3.3.2.1

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

$\frac{y \cdot - 2}{y \cdot y}$

Step 4.3.3.2.2

Cancel the common factor.

$\frac{y \cdot - 2}{y \cdot y}$

Step 4.3.3.2.3

Rewrite the expression.

$\frac{- 2}{y}$

Step 4.3.4

Substitute $y^{2}$ for $M$ .

$- \frac{2}{y}$

Step 4.4

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

$μ (x, y) = e^{\int - \frac{2}{y} d y}$

Step 5

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

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Step 5.1

Since $- 1$ is constant with respect to $y$ , move $- 1$ out of the integral.

$μ (x, y) = e^{- \int \frac{2}{y} d y}$

Step 5.2

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

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

Step 5.3

Multiply $2$ by $- 1$ .

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

Step 5.4

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

$μ (x, y) = e^{- 2 (\ln (| y |) + C)}$

Step 5.5

Simplify.

$μ (x, y) = e^{- 2 \ln (| y |)} + C$

Step 5.6

Simplify each term.

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Step 5.6.1

Simplify $- 2 \ln (| y |)$ by moving $- 2$ inside the logarithm.

$μ (x, y) = e^{\ln ({| y |}^{- 2})} + C$

Step 5.6.2

Exponentiation and log are inverse functions.

$μ (x, y) = {| y |}^{- 2} + C$

Step 5.6.3

Remove the absolute value in ${| y |}^{- 2}$ because exponentiations with even powers are always positive.

$μ (x, y) = y^{- 2} + C$

Step 5.6.4

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$μ (x, y) = \frac{1}{y^{2}} + C$

Step 6

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

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Step 6.1

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

$y^{2} \frac{1}{y^{2}} d t + (2 y t + 1) d y = 0$

Step 6.2

Cancel the common factor of $y^{2}$ .

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Step 6.2.1

Cancel the common factor.

$y^{2} \frac{1}{y^{2}} d t + (2 y t + 1) d y = 0$

Step 6.2.2

Rewrite the expression.

$1 d t + (2 y t + 1) d y = 0$

Step 6.3

Multiply $2 y t + 1$ by $\frac{1}{y^{2}}$ .

$1 d t + (2 y t + 1) \frac{1}{y^{2}} d y = 0$

Step 6.4

Multiply $2 y t + 1$ by $\frac{1}{y^{2}}$ .

$1 d t + \frac{2 y t + 1}{y^{2}} d y = 0$

Step 7

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

$f (x, y) = \int d x$

Step 8

Integrate $M (x, y) = 1$ to find $f (x, y)$ .

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Step 8.1

Apply the constant rule.

$f (x, y) = x + C$

Step 9

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

$f (x, y) = x + g (y)$

Step 10

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

$\frac{\partial f}{\partial y} = \frac{2 y t + 1}{y^{2}}$

Step 11

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

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Step 11.1

Differentiate $f$ with respect to $y$ .

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

Step 11.2

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

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

Step 11.3

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

$0 + \frac{d}{d y} [g (y)] = \frac{2 y t + 1}{y^{2}}$

Step 11.4

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

$0 + \frac{d g}{d y} = \frac{2 y t + 1}{y^{2}}$

Step 11.5

Add $0$ and $\frac{d g}{d y}$ .

$\frac{d g}{d y} = \frac{2 y t + 1}{y^{2}}$

Step 12

Find the antiderivative of $\frac{2 y t + 1}{y^{2}}$ to find $g (y)$ .

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Step 12.1

Integrate both sides of $\frac{d g}{d y} = \frac{2 y t + 1}{y^{2}}$ .

$\int \frac{d g}{d y} d y = \int \frac{2 y t + 1}{y^{2}} d y$

Step 12.2

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

$g (y) = \int \frac{2 y t + 1}{y^{2}} d y$

Step 12.3

Move $y^{2}$ out of the denominator by raising it to the $- 1$ power.

$g (y) = \int (2 y t + 1) {(y^{2})}^{- 1} d y$

Step 12.4

Multiply the exponents in ${(y^{2})}^{- 1}$ .

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Step 12.4.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$g (y) = \int (2 y t + 1) y^{2 \cdot - 1} d y$

Step 12.4.2

Multiply $2$ by $- 1$ .

$g (y) = \int (2 y t + 1) y^{- 2} d y$

Step 12.5

Expand $(2 y t + 1) y^{- 2}$ .

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Step 12.5.1

Apply the distributive property.

$g (y) = \int 2 y t y^{- 2} + 1 y^{- 2} d y$

Step 12.5.2

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

$g (y) = \int 2 y t y^{- 2} + y^{- 2} d y$

Step 12.6

Multiply $y$ by $y^{- 2}$ by adding the exponents.

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Step 12.6.1

Move $y^{- 2}$ .

$g (y) = \int 2 (y^{- 2} y) t + y^{- 2} d y$

Step 12.6.2

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

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Step 12.6.2.1

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

$g (y) = \int 2 (y^{- 2} y^{1}) t + y^{- 2} d y$

Step 12.6.2.2

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

$g (y) = \int 2 y^{- 2 + 1} t + y^{- 2} d y$

Step 12.6.3

Add $- 2$ and $1$ .

$g (y) = \int 2 y^{- 1} t + y^{- 2} d y$

Step 12.7

Split the single integral into multiple integrals.

$g (y) = \int 2 y^{- 1} t d y + \int y^{- 2} d y$

Step 12.8

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

$g (y) = 2 t \int y^{- 1} d y + \int y^{- 2} d y$

Step 12.9

The integral of $y^{- 1}$ with respect to $y$ is $\ln (| y |)$ .

$g (y) = 2 t (\ln (| y |) + C) + \int y^{- 2} d y$

Step 12.10

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

$g (y) = 2 t (\ln (| y |) + C) - y^{- 1} + C$

Step 12.11

Simplify.

$g (y) = 2 t \ln (| y |) - \frac{1}{y} + C$

Step 13

Substitute for $g (y)$ in $f (x, y) = x + g (y)$ .

$f (x, y) = x + 2 t \ln (| y |) - \frac{1}{y} + C$

Step 14

Simplify each term.

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Step 14.1

Simplify $2 \ln (| y |)$ by moving $2$ inside the logarithm.

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

Step 14.2

Remove the absolute value in ${| y |}^{2}$ because exponentiations with even powers are always positive.

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