Calculus Examples

$\sin (x) d y + y^{2} \cos (x) d x = 0$

Step 1

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

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Step 1.1

Rewrite.

$y^{2} \cos (x) d x + \sin (x) d y = 0$

Step 2

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

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Step 2.1

Differentiate $M$ with respect to $y$ .

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

Step 2.2

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

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

Step 2.3

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

$\frac{\partial M}{\partial y} = \cos (x) (2 y)$

Step 2.4

Reorder.

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Step 2.4.1

Move $2$ to the left of $\cos (x)$ .

$\frac{\partial M}{\partial y} = 2 \cdot \cos (x) y$

Step 2.4.2

Reorder the factors of $2 \cos (x) y$ .

$\frac{\partial M}{\partial y} = 2 y \cos (x)$

Step 3

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

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Step 3.1

Differentiate $N$ with respect to $x$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [\sin (x)]$

Step 3.2

The derivative of $\sin (x)$ with respect to $x$ is $\cos (x)$ .

$\frac{\partial N}{\partial x} = \cos (x)$

Step 4

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

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Step 4.1

Substitute $2 y \cos (x)$ for $\frac{\partial M}{\partial y}$ and $\cos (x)$ for $\frac{\partial N}{\partial x}$ .

$2 y \cos (x) = \cos (x)$

Step 4.2

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

$2 y \cos (x) = \cos (x)$ 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 $\cos (x)$ for $\frac{\partial N}{\partial x}$ .

$\frac{\cos (x) - \frac{\partial M}{\partial y}}{M}$

Step 5.2

Substitute $2 y \cos (x)$ for $\frac{\partial M}{\partial y}$ .

$\frac{\cos (x) - (2 y \cos (x))}{M}$

Step 5.3

Substitute $y^{2} \cos (x)$ for $M$ .

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Step 5.3.1

Substitute $y^{2} \cos (x)$ for $M$ .

$\frac{\cos (x) - (2 y \cos (x))}{y^{2} \cos (x)}$

Step 5.3.2

Simplify the numerator.

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Step 5.3.2.1

Factor $\cos (x)$ out of $\cos (x) - 1 \cdot 2 y \cos (x)$ .

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Step 5.3.2.1.1

Multiply by $1$ .

$\frac{\cos (x) \cdot 1 - 1 \cdot 2 y \cos (x)}{y^{2} \cos (x)}$

Step 5.3.2.1.2

Factor $\cos (x)$ out of $- 1 \cdot 2 y \cos (x)$ .

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

Step 5.3.2.1.3

Factor $\cos (x)$ out of $\cos (x) \cdot 1 + \cos (x) (- 1 \cdot 2 y)$ .

$\frac{\cos (x) (1 - 1 \cdot 2 y)}{y^{2} \cos (x)}$

Step 5.3.2.2

Multiply $- 1$ by $2$ .

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

Step 5.3.3

Substitute $y^{2} \cos (x)$ for $M$ .

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Step 5.3.3.1

Cancel the common factor.

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

Step 5.3.3.2

Rewrite the expression.

$\frac{1 - 2 y}{y^{2}}$

Step 5.4

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

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

Step 6

Evaluate the integral $e^{\int \frac{1 - 2 y}{y^{2}} d y}$ .

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Step 6.1

Apply basic rules of exponents.

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Step 6.1.1

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

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

Step 6.1.2

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

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Step 6.1.2.1

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

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

Step 6.1.2.2

Multiply $2$ by $- 1$ .

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

Step 6.2

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

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

Step 6.3

Simplify.

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Step 6.3.1

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

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

Step 6.3.2

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

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Step 6.3.2.1

Move $y^{- 2}$ .

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

Step 6.3.2.2

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

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Step 6.3.2.2.1

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

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

Step 6.3.2.2.2

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

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

Step 6.3.2.3

Add $- 2$ and $1$ .

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

Step 6.4

Split the single integral into multiple integrals.

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

Step 6.5

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

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

Step 6.6

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

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

Step 6.7

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

$μ (x, y) = e^{- y^{- 1} + C - 2 (\ln (| y |) + C)}$

Step 6.8

Simplify.

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

Step 6.9

Simplify each term.

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Step 6.9.1

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

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

Step 6.9.2

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

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

Step 7

Multiply both sides of $y^{2} \cos (x) d x + \sin (x) d y = 0$ by the integration factor $e^{- \frac{1}{y} - \ln (y^{2})}$ .

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Step 7.1

Multiply $y^{2} \cos (x)$ by $e^{- \frac{1}{y} - \ln (y^{2})}$ .

$y^{2} \cos (x) e^{- \frac{1}{y} - \ln (y^{2})} d x + \sin (x) d y = 0$

Step 7.2

Reorder factors in $y^{2} \cos (x) e^{- \frac{1}{y} - \ln (y^{2})}$ .

$y^{2} e^{- \frac{1}{y} - \ln (y^{2})} \cos (x) d x + \sin (x) d y = 0$

Step 7.3

Multiply $\sin (x)$ by $e^{- \frac{1}{y} - \ln (y^{2})}$ .

$y^{2} e^{- \frac{1}{y} - \ln (y^{2})} \cos (x) d x + \sin (x) e^{- \frac{1}{y} - \ln (y^{2})} d y = 0$

Step 7.4

Reorder factors in $\sin (x) e^{- \frac{1}{y} - \ln (y^{2})}$ .

$y^{2} e^{- \frac{1}{y} - \ln (y^{2})} \cos (x) d x + e^{- \frac{1}{y} - \ln (y^{2})} \sin (x) d y = 0$

Step 8

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

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

Step 9

Integrate $M (x, y) = y^{2} e^{- \frac{1}{y} - \ln (y^{2})} \cos (x)$ to find $f (x, y)$ .

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Step 9.1

Rewrite $y^{2} e^{- \frac{1}{y} - \ln (y^{2})} \cos (x)$ as $y^{2} e^{- \frac{1}{y} - (2 \ln (y))} \cos (x)$ .

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

Step 9.2

Since $y^{2} e^{- \frac{1}{y} - (2 \ln (y))}$ is constant with respect to $x$ , move $y^{2} e^{- \frac{1}{y} - (2 \ln (y))}$ out of the integral.

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

Step 9.3

Multiply $2$ by $- 1$ .

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

Step 9.4

The integral of $\cos (x)$ with respect to $x$ is $\sin (x)$ .

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

Step 9.5

Simplify.

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

Step 10

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

$f (x, y) = y^{2} e^{- \frac{1}{y} - 2 \ln (y)} \sin (x) + g (y)$

Step 11

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

$\frac{\partial f}{\partial y} = e^{- \frac{1}{y} - \ln (y^{2})} \sin (x)$

Step 12

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

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Step 12.1

Differentiate $f$ with respect to $y$ .

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

Step 12.2

By the Sum Rule, the derivative of $y^{2} e^{- \frac{1}{y} - 2 \ln (y)} \sin (x) + g (y)$ with respect to $y$ is $\frac{d}{d y} [y^{2} e^{- \frac{1}{y} - 2 \ln (y)} \sin (x)] + \frac{d}{d y} [g (y)]$ .

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

Step 12.3

Evaluate $\frac{d}{d y} [y^{2} e^{- \frac{1}{y} - 2 \ln (y)} \sin (x)]$ .

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Step 12.3.1

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

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

Step 12.3.2

Differentiate using the Product Rule which states that $\frac{d}{d y} [f (y) g (y)]$ is $f (y) \frac{d}{d y} [g (y)] + g (y) \frac{d}{d y} [f (y)]$ where $f (y) = y^{2}$ and $g (y) = e^{- \frac{1}{y} - 2 \ln (y)}$ .

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

Step 12.3.3

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 $g (y) = - \frac{1}{y} - 2 \ln (y)$ .

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Step 12.3.3.1

To apply the Chain Rule, set $u$ as $- \frac{1}{y} - 2 \ln (y)$ .

$\sin (x) (y^{2} (\frac{d}{d u} [e^{u}] \frac{d}{d y} [- \frac{1}{y} - 2 \ln (y)]) + e^{- \frac{1}{y} - 2 \ln (y)} \frac{d}{d y} [y^{2}]) + \frac{d}{d y} [g (y)] = e^{- \frac{1}{y} - \ln (y^{2})} \sin (x)$

Step 12.3.3.2

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

$\sin (x) (y^{2} (e^{u} \frac{d}{d y} [- \frac{1}{y} - 2 \ln (y)]) + e^{- \frac{1}{y} - 2 \ln (y)} \frac{d}{d y} [y^{2}]) + \frac{d}{d y} [g (y)] = e^{- \frac{1}{y} - \ln (y^{2})} \sin (x)$

Step 12.3.3.3

Replace all occurrences of $u$ with $- \frac{1}{y} - 2 \ln (y)$ .

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

Step 12.3.4

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

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

Step 12.3.5

Differentiate using the Product Rule which states that $\frac{d}{d y} [f (y) g (y)]$ is $f (y) \frac{d}{d y} [g (y)] + g (y) \frac{d}{d y} [f (y)]$ where $f (y) = - 1$ and $g (y) = \frac{1}{y}$ .

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

Step 12.3.6

Rewrite $\frac{1}{y}$ as $y^{- 1}$ .

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

Step 12.3.7

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

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

Step 12.3.8

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

$\sin (x) (y^{2} (e^{- \frac{1}{y} - 2 \ln (y)} (- - y^{- 2} + \frac{1}{y} \cdot 0 + \frac{d}{d y} [- 2 \ln (y)])) + e^{- \frac{1}{y} - 2 \ln (y)} \frac{d}{d y} [y^{2}]) + \frac{d}{d y} [g (y)] = e^{- \frac{1}{y} - \ln (y^{2})} \sin (x)$

Step 12.3.9

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

$\sin (x) (y^{2} (e^{- \frac{1}{y} - 2 \ln (y)} (- - y^{- 2} + \frac{1}{y} \cdot 0 - 2 \frac{d}{d y} [\ln (y)])) + e^{- \frac{1}{y} - 2 \ln (y)} \frac{d}{d y} [y^{2}]) + \frac{d}{d y} [g (y)] = e^{- \frac{1}{y} - \ln (y^{2})} \sin (x)$

Step 12.3.10

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

$\sin (x) (y^{2} (e^{- \frac{1}{y} - 2 \ln (y)} (- - y^{- 2} + \frac{1}{y} \cdot 0 - 2 \frac{1}{y})) + e^{- \frac{1}{y} - 2 \ln (y)} \frac{d}{d y} [y^{2}]) + \frac{d}{d y} [g (y)] = e^{- \frac{1}{y} - \ln (y^{2})} \sin (x)$

Step 12.3.11

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

$\sin (x) (y^{2} (e^{- \frac{1}{y} - 2 \ln (y)} (- - y^{- 2} + \frac{1}{y} \cdot 0 - 2 \frac{1}{y})) + e^{- \frac{1}{y} - 2 \ln (y)} (2 y)) + \frac{d}{d y} [g (y)] = e^{- \frac{1}{y} - \ln (y^{2})} \sin (x)$

Step 12.3.12

Multiply $- 1$ by $- 1$ .

$\sin (x) (y^{2} (e^{- \frac{1}{y} - 2 \ln (y)} (1 y^{- 2} + \frac{1}{y} \cdot 0 - 2 \frac{1}{y})) + e^{- \frac{1}{y} - 2 \ln (y)} (2 y)) + \frac{d}{d y} [g (y)] = e^{- \frac{1}{y} - \ln (y^{2})} \sin (x)$

Step 12.3.13

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

$\sin (x) (y^{2} (e^{- \frac{1}{y} - 2 \ln (y)} (y^{- 2} + \frac{1}{y} \cdot 0 - 2 \frac{1}{y})) + e^{- \frac{1}{y} - 2 \ln (y)} (2 y)) + \frac{d}{d y} [g (y)] = e^{- \frac{1}{y} - \ln (y^{2})} \sin (x)$

Step 12.3.14

Multiply $\frac{1}{y}$ by $0$ .

$\sin (x) (y^{2} (e^{- \frac{1}{y} - 2 \ln (y)} (y^{- 2} + 0 - 2 \frac{1}{y})) + e^{- \frac{1}{y} - 2 \ln (y)} (2 y)) + \frac{d}{d y} [g (y)] = e^{- \frac{1}{y} - \ln (y^{2})} \sin (x)$

Step 12.3.15

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

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

Step 12.3.16

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

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

Step 12.3.17

Move the negative in front of the fraction.

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

Step 12.4

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

$\sin (x) (y^{2} (e^{- \frac{1}{y} - 2 \ln (y)} (y^{- 2} - \frac{2}{y})) + e^{- \frac{1}{y} - 2 \ln (y)} (2 y)) + \frac{d g}{d y} = e^{- \frac{1}{y} - \ln (y^{2})} \sin (x)$

Step 12.5

Simplify.

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Step 12.5.1

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

$\sin (x) (y^{2} (e^{- \frac{1}{y} - 2 \ln (y)} (\frac{1}{y^{2}} - \frac{2}{y})) + e^{- \frac{1}{y} - 2 \ln (y)} (2 y)) + \frac{d g}{d y} = e^{- \frac{1}{y} - \ln (y^{2})} \sin (x)$

Step 12.5.2

Apply the distributive property.

$\sin (x) (y^{2} (e^{- \frac{1}{y} - 2 \ln (y)} (\frac{1}{y^{2}} - \frac{2}{y}))) + \sin (x) (e^{- \frac{1}{y} - 2 \ln (y)} (2 y)) + \frac{d g}{d y} = e^{- \frac{1}{y} - \ln (y^{2})} \sin (x)$

Step 12.5.3

Remove parentheses.

$\sin (x) y^{2} (e^{- \frac{1}{y} - 2 \ln (y)} (\frac{1}{y^{2}} - \frac{2}{y})) + \sin (x) e^{- \frac{1}{y} - 2 \ln (y)} (2 y) + \frac{d g}{d y} = e^{- \frac{1}{y} - \ln (y^{2})} \sin (x)$

Step 12.5.4

Reorder terms.

$\frac{d g}{d y} + e^{- \frac{1}{y} - 2 \ln (y)} y^{2} \sin (x) (\frac{1}{y^{2}} - \frac{2}{y}) + 2 e^{- \frac{1}{y} - 2 \ln (y)} y \sin (x) = e^{- \frac{1}{y} - \ln (y^{2})} \sin (x)$

Step 12.5.5

Simplify each term.

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Step 12.5.5.1

Apply the distributive property.

$\frac{d g}{d y} + e^{- \frac{1}{y} - 2 \ln (y)} y^{2} \sin (x) \frac{1}{y^{2}} + e^{- \frac{1}{y} - 2 \ln (y)} y^{2} \sin (x) (- \frac{2}{y}) + 2 e^{- \frac{1}{y} - 2 \ln (y)} y \sin (x) = e^{- \frac{1}{y} - \ln (y^{2})} \sin (x)$

Step 12.5.5.2

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

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Step 12.5.5.2.1

Factor $y^{2}$ out of $e^{- \frac{1}{y} - 2 \ln (y)} y^{2} \sin (x)$ .

$\frac{d g}{d y} + y^{2} (e^{- \frac{1}{y} - 2 \ln (y)} \sin (x)) \frac{1}{y^{2}} + e^{- \frac{1}{y} - 2 \ln (y)} y^{2} \sin (x) (- \frac{2}{y}) + 2 e^{- \frac{1}{y} - 2 \ln (y)} y \sin (x) = e^{- \frac{1}{y} - \ln (y^{2})} \sin (x)$

Step 12.5.5.2.2

Cancel the common factor.

Step 12.5.5.2.3

Rewrite the expression.

$\frac{d g}{d y} + e^{- \frac{1}{y} - 2 \ln (y)} \sin (x) + e^{- \frac{1}{y} - 2 \ln (y)} y^{2} \sin (x) (- \frac{2}{y}) + 2 e^{- \frac{1}{y} - 2 \ln (y)} y \sin (x) = e^{- \frac{1}{y} - \ln (y^{2})} \sin (x)$

Step 12.5.5.3

Rewrite using the commutative property of multiplication.

$\frac{d g}{d y} + e^{- \frac{1}{y} - 2 \ln (y)} \sin (x) - e^{- \frac{1}{y} - 2 \ln (y)} y^{2} \sin (x) \frac{2}{y} + 2 e^{- \frac{1}{y} - 2 \ln (y)} y \sin (x) = e^{- \frac{1}{y} - \ln (y^{2})} \sin (x)$

Step 12.5.5.4

Simplify each term.

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Step 12.5.5.4.1

Cancel the common factor of $y$ .

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Step 12.5.5.4.1.1

Factor $y$ out of $- e^{- \frac{1}{y} - 2 \ln (y)} y^{2} \sin (x)$ .

$\frac{d g}{d y} + e^{- \frac{1}{y} - 2 \ln (y)} \sin (x) + y (- e^{- \frac{1}{y} - 2 \ln (y)} y \sin (x)) \frac{2}{y} + 2 e^{- \frac{1}{y} - 2 \ln (y)} y \sin (x) = e^{- \frac{1}{y} - \ln (y^{2})} \sin (x)$

Step 12.5.5.4.1.2

Cancel the common factor.

Step 12.5.5.4.1.3

Rewrite the expression.

$\frac{d g}{d y} + e^{- \frac{1}{y} - 2 \ln (y)} \sin (x) - e^{- \frac{1}{y} - 2 \ln (y)} y \sin (x) \cdot 2 + 2 e^{- \frac{1}{y} - 2 \ln (y)} y \sin (x) = e^{- \frac{1}{y} - \ln (y^{2})} \sin (x)$

Step 12.5.5.4.2

Multiply $2$ by $- 1$ .

$\frac{d g}{d y} + e^{- \frac{1}{y} - 2 \ln (y)} \sin (x) - 2 e^{- \frac{1}{y} - 2 \ln (y)} y \sin (x) + 2 e^{- \frac{1}{y} - 2 \ln (y)} y \sin (x) = e^{- \frac{1}{y} - \ln (y^{2})} \sin (x)$

Step 12.5.6

Combine the opposite terms in $\frac{d g}{d y} + e^{- \frac{1}{y} - 2 \ln (y)} \sin (x) - 2 e^{- \frac{1}{y} - 2 \ln (y)} y \sin (x) + 2 e^{- \frac{1}{y} - 2 \ln (y)} y \sin (x)$ .

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Step 12.5.6.1

Add $- 2 e^{- \frac{1}{y} - 2 \ln (y)} y \sin (x)$ and $2 e^{- \frac{1}{y} - 2 \ln (y)} y \sin (x)$ .

$\frac{d g}{d y} + e^{- \frac{1}{y} - 2 \ln (y)} \sin (x) + 0 = e^{- \frac{1}{y} - \ln (y^{2})} \sin (x)$

Step 12.5.6.2

Add $\frac{d g}{d y} + e^{- \frac{1}{y} - 2 \ln (y)} \sin (x)$ and $0$ .

$\frac{d g}{d y} + e^{- \frac{1}{y} - 2 \ln (y)} \sin (x) = e^{- \frac{1}{y} - \ln (y^{2})} \sin (x)$

Step 13

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

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Step 13.1

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

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Step 13.1.1

Move all the terms containing a logarithm to the left side of the equation.

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

Step 13.1.2

Simplify the left side.

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Step 13.1.2.1

Simplify $e^{- \frac{1}{y} - 2 \ln (y)} \sin (x) - e^{- \frac{1}{y} - \ln (y^{2})} \sin (x)$ .

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Step 13.1.2.1.1

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

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

Step 13.1.2.1.2

Subtract $e^{- \frac{1}{y} - \ln (y^{2})} \sin (x)$ from $e^{- \frac{1}{y} - \ln (y^{2})} \sin (x)$ .

$0 = - \frac{d g}{d y}$

Step 13.1.3

Since $\frac{d g}{d y}$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$- \frac{d g}{d y} = 0$

Step 13.1.4

Divide each term in $- \frac{d g}{d y} = 0$ by $- 1$ and simplify.

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Step 13.1.4.1

Divide each term in $- \frac{d g}{d y} = 0$ by $- 1$ .

$\frac{- \frac{d g}{d y}}{- 1} = \frac{0}{- 1}$

Step 13.1.4.2

Simplify the left side.

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Step 13.1.4.2.1

Dividing two negative values results in a positive value.

$\frac{\frac{d g}{d y}}{1} = \frac{0}{- 1}$

Step 13.1.4.2.2

Divide $\frac{d g}{d y}$ by $1$ .

$\frac{d g}{d y} = \frac{0}{- 1}$

Step 13.1.4.3

Simplify the right side.

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Step 13.1.4.3.1

Divide $0$ by $- 1$ .

$\frac{d g}{d y} = 0$

Step 14

Find the antiderivative of $0$ to find $g (y)$ .

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Step 14.1

Integrate both sides of $\frac{d g}{d y} = 0$ .

$\int \frac{d g}{d y} d y = \int 0 d y$

Step 14.2

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

$g (y) = \int 0 d y$

Step 14.3

The integral of $0$ with respect to $y$ is $0$ .

$g (y) = 0 + C$

Step 14.4

Add $0$ and $C$ .

$g (y) = C$

Step 15

Substitute for $g (y)$ in $f (x, y) = y^{2} e^{- \frac{1}{y} - 2 \ln (y)} \sin (x) + g (y)$ .

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

Step 16

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

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