Calculus Examples

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

Step 1

Find

$\frac{\partial M}{\partial y}$ where

$M (x, y) = 2 x y$ .

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Step 1.1

Differentiate

$M$ with respect to

$y$ .

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

Step 1.2

Since

$2 x$ is constant with respect to

$y$ , the derivative of

$2 x y$ with respect to

$y$ is

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

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

Step 1.3

Differentiate using the Power Rule which states that

$\frac{d}{d y} [y^{n}]$ is

$n y^{n - 1}$ where

$n = 1$ .

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

Step 1.4

Multiply

$2$ by

$1$ .

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

Step 2

Find

$\frac{\partial N}{\partial x}$ where

$N (x, y) = y^{2} - 3 x^{2}$ .

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Step 2.1

Differentiate

$N$ with respect to

$x$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [y^{2} - 3 x^{2}]$

Step 2.2

Differentiate.

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Step 2.2.1

By the Sum Rule, the derivative of

$y^{2} - 3 x^{2}$ with respect to

$x$ is

$\frac{d}{d x} [y^{2}] + \frac{d}{d x} [- 3 x^{2}]$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [y^{2}] + \frac{d}{d x} [- 3 x^{2}]$

Step 2.2.2

Since

$y^{2}$ is constant with respect to

$x$ , the derivative of

$y^{2}$ with respect to

$x$ is

$0$ .

$\frac{\partial N}{\partial x} = 0 + \frac{d}{d x} [- 3 x^{2}]$

Step 2.3

Evaluate

$\frac{d}{d x} [- 3 x^{2}]$ .

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Step 2.3.1

Since

$- 3$ is constant with respect to

$x$ , the derivative of

$- 3 x^{2}$ with respect to

$x$ is

$- 3 \frac{d}{d x} [x^{2}]$ .

$\frac{\partial N}{\partial x} = 0 - 3 \frac{d}{d x} [x^{2}]$

Step 2.3.2

Differentiate using the Power Rule which states that

$\frac{d}{d x} [x^{n}]$ is

$n x^{n - 1}$ where

$n = 2$ .

$\frac{\partial N}{\partial x} = 0 - 3 (2 x)$

Step 2.3.3

Multiply

$2$ by

$- 3$ .

$\frac{\partial N}{\partial x} = 0 - 6 x$

Step 2.4

Subtract

$6 x$ from

$0$ .

$\frac{\partial N}{\partial x} = - 6 x$

Step 3

Check that

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

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Step 3.1

Substitute

$2 x$ for

$\frac{\partial M}{\partial y}$ and

$- 6 x$ for

$\frac{\partial N}{\partial x}$ .

$2 x = - 6 x$

Step 3.2

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

$2 x = - 6 x$ 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

$- 6 x$ for

$\frac{\partial N}{\partial x}$ .

$\frac{- 6 x - \frac{\partial M}{\partial y}}{M}$

Step 4.2

Substitute

$2 x$ for

$\frac{\partial M}{\partial y}$ .

$\frac{- 6 x - (2 x)}{M}$

Step 4.3

Substitute

$2 x y$ for

$M$ .

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Step 4.3.1

Substitute

$2 x y$ for

$M$ .

$\frac{- 6 x - (2 x)}{2 x y}$

Step 4.3.2

Simplify the numerator.

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Step 4.3.2.1

Factor

$x$ out of

$- 6 x - 1 \cdot 2 x$ .

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Step 4.3.2.1.1

Factor

$x$ out of

$- 6 x$ .

$\frac{x \cdot - 6 - 1 \cdot 2 x}{2 x y}$

Step 4.3.2.1.2

Factor

$x$ out of

$- 1 \cdot 2 x$ .

$\frac{x \cdot - 6 + x (- 1 \cdot 2)}{2 x y}$

Step 4.3.2.1.3

Factor

$x$ out of

$x \cdot - 6 + x (- 1 \cdot 2)$ .

$\frac{x (- 6 - 1 \cdot 2)}{2 x y}$

Step 4.3.2.2

Multiply

$- 1$ by

$2$ .

$\frac{x (- 6 - 2)}{2 x y}$

Step 4.3.2.3

Subtract

$2$ from

$- 6$ .

$\frac{x \cdot - 8}{2 x y}$

Step 4.3.3

Cancel the common factor of

$x$ .

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Step 4.3.3.1

Cancel the common factor.

$\frac{x \cdot - 8}{2 x y}$

Step 4.3.3.2

Rewrite the expression.

$\frac{- 8}{2 y}$

Step 4.3.4

Cancel the common factor of

$- 8$ and

$2$ .

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Step 4.3.4.1

Factor

$2$ out of

$- 8$ .

$\frac{2 \cdot - 4}{2 y}$

Step 4.3.4.2

Cancel the common factors.

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Step 4.3.4.2.1

Factor

$2$ out of

$2 y$ .

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

Step 4.3.4.2.2

Cancel the common factor.

$\frac{2 \cdot - 4}{2 y}$

Step 4.3.4.2.3

Rewrite the expression.

$\frac{- 4}{y}$

Step 4.3.5

Substitute

$2 x y$ for

$M$ .

$- \frac{4}{y}$

Step 4.4

Find the integration factor

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

$μ (x, y) = e^{\int - \frac{4}{y} d y}$

Step 5

Evaluate the integral

$e^{\int - \frac{4}{y} d y}$ .

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Step 5.1

Since

$- 1$ is constant with respect to

$y$ , move

$- 1$ out of the integral.

$μ (x, y) = e^{- \int \frac{4}{y} d y}$

Step 5.2

Since

$4$ is constant with respect to

$y$ , move

$4$ out of the integral.

$μ (x, y) = e^{- (4 \int \frac{1}{y} d y)}$

Step 5.3

Multiply

$4$ by

$- 1$ .

$μ (x, y) = e^{- 4 \int \frac{1}{y} d y}$

Step 5.4

The integral of

$\frac{1}{y}$ with respect to

$y$ is

$\ln (| y |)$ .

$μ (x, y) = e^{- 4 (\ln (| y |) + C)}$

Step 5.5

Simplify.

$μ (x, y) = e^{- 4 \ln (| y |)} + C$

Step 5.6

Simplify each term.

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Step 5.6.1

Simplify

$- 4 \ln (| y |)$ by moving

$- 4$ inside the logarithm.

$μ (x, y) = e^{\ln ({| y |}^{- 4})} + C$

Step 5.6.2

Exponentiation and log are inverse functions.

$μ (x, y) = {| y |}^{- 4} + C$

Step 5.6.3

Remove the absolute value in

${| y |}^{- 4}$ because exponentiations with even powers are always positive.

$μ (x, y) = y^{- 4} + C$

Step 5.6.4

Rewrite the expression using the negative exponent rule

$b^{- n} = \frac{1}{b^{n}}$ .

$μ (x, y) = \frac{1}{y^{4}} + C$

Step 6

Multiply both sides of

$2 x y d x + (y^{2} - 3 x^{2}) d y = 0$ by the integration factor

$\frac{1}{y^{4}}$ .

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Step 6.1

Multiply

$2 x y$ by

$\frac{1}{y^{4}}$ .

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

Step 6.2

Cancel the common factor of

$y$ .

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Step 6.2.1

Factor

$y$ out of

$2 x y$ .

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

Step 6.2.2

Factor

$y$ out of

$y^{4}$ .

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

Step 6.2.3

Cancel the common factor.

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

Step 6.2.4

Rewrite the expression.

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

Step 6.3

Combine

$2$ and

$\frac{1}{y^{3}}$ .

$x \frac{2}{y^{3}} d x + (y^{2} - 3 x^{2}) d y = 0$

Step 6.4

Combine

$x$ and

$\frac{2}{y^{3}}$ .

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

Step 6.5

Move

$2$ to the left of

$x$ .

$\frac{2 x}{y^{3}} d x + (y^{2} - 3 x^{2}) d y = 0$

Step 6.6

Multiply

$y^{2} - 3 x^{2}$ by

$\frac{1}{y^{4}}$ .

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

Step 6.7

Multiply

$y^{2} - 3 x^{2}$ by

$\frac{1}{y^{4}}$ .

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

Step 7

Set

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

$M (x, y)$ .

$f (x, y) = \int \frac{2 x}{y^{3}} d x$

Step 8

Integrate

$M (x, y) = \frac{2 x}{y^{3}}$ to find

$f (x, y)$ .

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Step 8.1

Since

$\frac{2}{y^{3}}$ is constant with respect to

$x$ , move

$\frac{2}{y^{3}}$ out of the integral.

$f (x, y) = \frac{2}{y^{3}} \int x d x$

Step 8.2

By the Power Rule, the integral of

$x$ with respect to

$x$ is

$\frac{1}{2} x^{2}$ .

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

Step 8.3

Simplify the answer.

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Step 8.3.1

Rewrite

$\frac{2}{y^{3}} (\frac{1}{2} x^{2} + C)$ as

$\frac{2}{y^{3}} \cdot \frac{1}{2} x^{2} + C$ .

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

Step 8.3.2

Simplify.

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Step 8.3.2.1

Multiply

$\frac{2}{y^{3}}$ by

$\frac{1}{2}$ .

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

Step 8.3.2.2

Move

$2$ to the left of

$y^{3}$ .

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

Step 8.3.2.3

Multiply

$2$ by

$y^{3}$ .

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

Step 8.3.2.4

Cancel the common factor of

$2$ .

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Step 8.3.2.4.1

Cancel the common factor.

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

Step 8.3.2.4.2

Rewrite the expression.

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

Step 8.3.2.5

Combine

$\frac{1}{y^{3}}$ and

$x^{2}$ .

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

Step 9

Since the integral of

$g (y)$ will contain an integration constant, we can replace

$C$ with

$g (y)$ .

$f (x, y) = \frac{x^{2}}{y^{3}} + g (y)$

Step 10

Set

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

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

Step 11

Find

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

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Step 11.1

Differentiate

$f$ with respect to

$y$ .

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

Step 11.2

By the Sum Rule, the derivative of

$\frac{x^{2}}{y^{3}} + g (y)$ with respect to

$y$ is

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

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

Step 11.3

Evaluate

$\frac{d}{d y} [\frac{x^{2}}{y^{3}}]$ .

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Step 11.3.1

Since

$x^{2}$ is constant with respect to

$y$ , the derivative of

$\frac{x^{2}}{y^{3}}$ with respect to

$y$ is

$x^{2} \frac{d}{d y} [\frac{1}{y^{3}}]$ .

$x^{2} \frac{d}{d y} [\frac{1}{y^{3}}] + \frac{d}{d y} [g (y)] = \frac{y^{2} - 3 x^{2}}{y^{4}}$

Step 11.3.2

Rewrite

$\frac{1}{y^{3}}$ as

${(y^{3})}^{- 1}$ .

$x^{2} \frac{d}{d y} [{(y^{3})}^{- 1}] + \frac{d}{d y} [g (y)] = \frac{y^{2} - 3 x^{2}}{y^{4}}$

Step 11.3.3

Differentiate using the chain rule, which states that

$\frac{d}{d y} [f (g (y))]$ is

$f' (g (y)) g' (y)$ where

$f (y) = y^{- 1}$ and

$g (y) = y^{3}$ .

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Step 11.3.3.1

To apply the Chain Rule, set

$u$ as

$y^{3}$ .

$x^{2} (\frac{d}{d u} [u^{- 1}] \frac{d}{d y} [y^{3}]) + \frac{d}{d y} [g (y)] = \frac{y^{2} - 3 x^{2}}{y^{4}}$

Step 11.3.3.2

Differentiate using the Power Rule which states that

$\frac{d}{d u} [u^{n}]$ is

$n u^{n - 1}$ where

$n = - 1$ .

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

Step 11.3.3.3

Replace all occurrences of

$u$ with

$y^{3}$ .

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

Step 11.3.4

Differentiate using the Power Rule which states that

$\frac{d}{d y} [y^{n}]$ is

$n y^{n - 1}$ where

$n = 3$ .

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

Step 11.3.5

Multiply the exponents in

${(y^{3})}^{- 2}$ .

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Step 11.3.5.1

Apply the power rule and multiply exponents,

${(a^{m})}^{n} = a^{m n}$ .

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

Step 11.3.5.2

Multiply

$3$ by

$- 2$ .

$x^{2} (- y^{- 6} (3 y^{2})) + \frac{d}{d y} [g (y)] = \frac{y^{2} - 3 x^{2}}{y^{4}}$

Step 11.3.6

Multiply

$3$ by

$- 1$ .

$x^{2} (- 3 y^{- 6} y^{2}) + \frac{d}{d y} [g (y)] = \frac{y^{2} - 3 x^{2}}{y^{4}}$

Step 11.3.7

Multiply

$y^{- 6}$ by

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

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Step 11.3.7.1

Move

$y^{2}$ .

$x^{2} (- 3 (y^{2} y^{- 6})) + \frac{d}{d y} [g (y)] = \frac{y^{2} - 3 x^{2}}{y^{4}}$

Step 11.3.7.2

Use the power rule

$a^{m} a^{n} = a^{m + n}$ to combine exponents.

$x^{2} (- 3 y^{2 - 6}) + \frac{d}{d y} [g (y)] = \frac{y^{2} - 3 x^{2}}{y^{4}}$

Step 11.3.7.3

Subtract

$6$ from

$2$ .

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

Step 11.4

Differentiate using the function rule which states that the derivative of

$g (y)$ is

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

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

Step 11.5

Simplify.

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Step 11.5.1

Rewrite the expression using the negative exponent rule

$b^{- n} = \frac{1}{b^{n}}$ .

$x^{2} (- 3 \frac{1}{y^{4}}) + \frac{d g}{d y} = \frac{y^{2} - 3 x^{2}}{y^{4}}$

Step 11.5.2

Combine terms.

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Step 11.5.2.1

Combine

$- 3$ and

$\frac{1}{y^{4}}$ .

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

Step 11.5.2.2

Move the negative in front of the fraction.

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

Step 11.5.2.3

Combine

$x^{2}$ and

$\frac{3}{y^{4}}$ .

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

Step 11.5.2.4

Move

$3$ to the left of

$x^{2}$ .

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

Step 11.5.3

Reorder terms.

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

Step 12

Solve for

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

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Step 12.1

Solve for

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

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Step 12.1.1

Move all terms containing variables to the left side of the equation.

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Step 12.1.1.1

Subtract

$\frac{y^{2} - 3 x^{2}}{y^{4}}$ from both sides of the equation.

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

Step 12.1.1.2

Combine the numerators over the common denominator.

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

Step 12.1.1.3

Simplify each term.

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Step 12.1.1.3.1

Apply the distributive property.

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

Step 12.1.1.3.2

Multiply

$- 3$ by

$- 1$ .

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

Step 12.1.1.4

Combine the opposite terms in

$- 3 x^{2} - y^{2} + 3 x^{2}$ .

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Step 12.1.1.4.1

Add

$- 3 x^{2}$ and

$3 x^{2}$ .

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

Step 12.1.1.4.2

Add

$- y^{2}$ and

$0$ .

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

Step 12.1.1.5

Simplify each term.

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Step 12.1.1.5.1

Cancel the common factor of

$y^{2}$ and

$y^{4}$ .

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Step 12.1.1.5.1.1

Factor

$y^{2}$ out of

$- y^{2}$ .

$\frac{d g}{d y} + \frac{y^{2} \cdot - 1}{y^{4}} = 0$

Step 12.1.1.5.1.2

Cancel the common factors.

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Step 12.1.1.5.1.2.1

Factor

$y^{2}$ out of

$y^{4}$ .

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

Step 12.1.1.5.1.2.2

Cancel the common factor.

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

Step 12.1.1.5.1.2.3

Rewrite the expression.

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

Step 12.1.1.5.2

Move the negative in front of the fraction.

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

Step 12.1.2

Add

$\frac{1}{y^{2}}$ to both sides of the equation.

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

Step 13

Find the antiderivative of

$\frac{1}{y^{2}}$ to find

$g (y)$ .

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Step 13.1

Integrate both sides of

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

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

Step 13.2

Evaluate

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

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

Step 13.3

Move

$y^{2}$ out of the denominator by raising it to the

$- 1$ power.

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

Step 13.4

Multiply the exponents in

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

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Step 13.4.1

Apply the power rule and multiply exponents,

${(a^{m})}^{n} = a^{m n}$ .

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

Step 13.4.2

Multiply

$2$ by

$- 1$ .

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

Step 13.5

By the Power Rule, the integral of

$y^{- 2}$ with respect to

$y$ is

$- y^{- 1}$ .

$g (y) = - y^{- 1} + C$

Step 13.6

Rewrite

$- y^{- 1} + C$ as

$- \frac{1}{y} + C$ .

$g (y) = - \frac{1}{y} + C$

Step 14

Substitute for

$g (y)$ in

$f (x, y) = \frac{x^{2}}{y^{3}} + g (y)$ .

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