Calculus Examples

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

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

Step 1

Find

\frac{\partial M}{\partial y}

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

M (x, y) = 3 x^{2} y + y^{2}

$M (x, y) = 3 x^{2} y + y^{2}$ .

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Step 1.1

Differentiate

M

$M$ with respect to

y

$y$ .

\frac{\partial M}{\partial y} = \frac{d}{d y} [3 x^{2} y + y^{2}]

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

Step 1.2

By the Sum Rule, the derivative of

3 x^{2} y + y^{2}

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

y

$y$ is

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

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

\frac{\partial M}{\partial y} = \frac{d}{d y} [3 x^{2} y] + \frac{d}{d y} [y^{2}]

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

Step 1.3

Evaluate

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

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

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Step 1.3.1

Since

3 x^{2}

$3 x^{2}$ is constant with respect to

y

$y$ , the derivative of

3 x^{2} y

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

y

$y$ is

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

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

\frac{\partial M}{\partial y} = 3 x^{2} \frac{d}{d y} [y] + \frac{d}{d y} [y^{2}]

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

Step 1.3.2

Differentiate using the Power Rule which states that

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

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

n y^{n - 1}

$n y^{n - 1}$ where

n = 1

$n = 1$ .

\frac{\partial M}{\partial y} = 3 x^{2} \cdot 1 + \frac{d}{d y} [y^{2}]

$\frac{\partial M}{\partial y} = 3 x^{2} \cdot 1 + \frac{d}{d y} [y^{2}]$

Step 1.3.3

Multiply

3

$3$ by

1

$1$ .

\frac{\partial M}{\partial y} = 3 x^{2} + \frac{d}{d y} [y^{2}]

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

\frac{\partial M}{\partial y} = 3 x^{2} + \frac{d}{d y} [y^{2}]

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

Step 1.4

Differentiate using the Power Rule which states that

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

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

n y^{n - 1}

$n y^{n - 1}$ where

n = 2

$n = 2$ .

\frac{\partial M}{\partial y} = 3 x^{2} + 2 y

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

\frac{\partial M}{\partial y} = 3 x^{2} + 2 y

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

Step 2

Find

\frac{\partial N}{\partial x}

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

N (x, y) = x^{3} + 2 x y + 3

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

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Step 2.1

Differentiate

N

$N$ with respect to

x

$x$ .

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

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

Step 2.2

Differentiate.

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Step 2.2.1

By the Sum Rule, the derivative of

x^{3} + 2 x y + 3

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

x

$x$ is

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

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

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

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

Step 2.2.2

Differentiate using the Power Rule which states that

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

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

n x^{n - 1}

$n x^{n - 1}$ where

n = 3

$n = 3$ .

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

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

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

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

Step 2.3

Evaluate

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

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

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Step 2.3.1

Since

2 y

$2 y$ is constant with respect to

x

$x$ , the derivative of

2 x y

$2 x y$ with respect to

x

$x$ is

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

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

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

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

Step 2.3.2

Differentiate using the Power Rule which states that

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

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

n x^{n - 1}

$n x^{n - 1}$ where

n = 1

$n = 1$ .

\frac{\partial N}{\partial x} = 3 x^{2} + 2 y \cdot 1 + \frac{d}{d x} [3]

$\frac{\partial N}{\partial x} = 3 x^{2} + 2 y \cdot 1 + \frac{d}{d x} [3]$

Step 2.3.3

Multiply

2

$2$ by

1

$1$ .

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

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

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

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

Step 2.4

Differentiate using the Constant Rule.

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Step 2.4.1

Since

3

$3$ is constant with respect to

x

$x$ , the derivative of

3

$3$ with respect to

x

$x$ is

0

$0$ .

\frac{\partial N}{\partial x} = 3 x^{2} + 2 y + 0

$\frac{\partial N}{\partial x} = 3 x^{2} + 2 y + 0$

Step 2.4.2

Add

3 x^{2} + 2 y

$3 x^{2} + 2 y$ and

0

$0$ .

\frac{\partial N}{\partial x} = 3 x^{2} + 2 y

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

\frac{\partial N}{\partial x} = 3 x^{2} + 2 y

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

\frac{\partial N}{\partial x} = 3 x^{2} + 2 y

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

Step 3

Check that

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

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

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Step 3.1

Substitute

3 x^{2} + 2 y

$3 x^{2} + 2 y$ for

\frac{\partial M}{\partial y}

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

3 x^{2} + 2 y

$3 x^{2} + 2 y$ for

\frac{\partial N}{\partial x}

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

3 x^{2} + 2 y = 3 x^{2} + 2 y

$3 x^{2} + 2 y = 3 x^{2} + 2 y$

Step 3.2

Since the two sides have been shown to be equivalent, the equation is an identity.

3 x^{2} + 2 y = 3 x^{2} + 2 y

$3 x^{2} + 2 y = 3 x^{2} + 2 y$ is an identity.

3 x^{2} + 2 y = 3 x^{2} + 2 y

$3 x^{2} + 2 y = 3 x^{2} + 2 y$ is an identity.

Step 4

Set

f (x, y)

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

M (x, y)

$M (x, y)$ .

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

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

Step 5

Integrate

M (x, y) = 3 x^{2} y + y^{2}

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

f (x, y)

$f (x, y)$ .

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Step 5.1

Split the single integral into multiple integrals.

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

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

Step 5.2

Since

3 y

$3 y$ is constant with respect to

x

$x$ , move

3 y

$3 y$ out of the integral.

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

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

Step 5.3

By the Power Rule, the integral of

x^{2}

$x^{2}$ with respect to

x

$x$ is

\frac{1}{3} x^{3}

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

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

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

Step 5.4

Apply the constant rule.

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

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

Step 5.5

Combine

\frac{1}{3}

$\frac{1}{3}$ and

x^{3}

$x^{3}$ .

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

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

Step 5.6

Simplify.

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

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

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

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

Step 6

Since the integral of

g (y)

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

C

$C$ with

g (y)

$g (y)$ .

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

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

Step 7

Set

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

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

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

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

Step 8

Find

\frac{\partial f}{\partial y}

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

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Step 8.1

Differentiate

f

$f$ with respect to

y

$y$ .

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

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

Step 8.2

By the Sum Rule, the derivative of

y x^{3} + y^{2} x + g (y)

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

y

$y$ is

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

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

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

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

Step 8.3

Evaluate

\frac{d}{d y} [y x^{3}]

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

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Step 8.3.1

Since

x^{3}

$x^{3}$ is constant with respect to

y

$y$ , the derivative of

y x^{3}

$y x^{3}$ with respect to

y

$y$ is

x^{3} \frac{d}{d y} [y]

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

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

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

Step 8.3.2

Differentiate using the Power Rule which states that

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

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

n y^{n - 1}

$n y^{n - 1}$ where

n = 1

$n = 1$ .

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

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

Step 8.3.3

Multiply

x^{3}

$x^{3}$ by

1

$1$ .

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

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

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

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

Step 8.4

Evaluate

\frac{d}{d y} [y^{2} x]

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

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Step 8.4.1

Since

x

$x$ is constant with respect to

y

$y$ , the derivative of

y^{2} x

$y^{2} x$ with respect to

y

$y$ is

x \frac{d}{d y} [y^{2}]

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

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

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

Step 8.4.2

Differentiate using the Power Rule which states that

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

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

n y^{n - 1}

$n y^{n - 1}$ where

n = 2

$n = 2$ .

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

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

Step 8.4.3

Move

2

$2$ to the left of

x

$x$ .

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

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

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

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

Step 8.5

Differentiate using the function rule which states that the derivative of

g (y)

$g (y)$ is

\frac{d g}{d y}

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

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

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

Step 8.6

Reorder terms.

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

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

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

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

Step 9

Solve for

\frac{d g}{d y}

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

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Step 9.1

Move all terms not containing

\frac{d g}{d y}

$\frac{d g}{d y}$ to the right side of the equation.

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Step 9.1.1

Subtract

x^{3}

$x^{3}$ from both sides of the equation.

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

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

Step 9.1.2

Subtract

2 x y

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

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

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

Step 9.1.3

Combine the opposite terms in

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

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

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Step 9.1.3.1

Subtract

x^{3}

$x^{3}$ from

x^{3}

$x^{3}$ .

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

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

Step 9.1.3.2

Add

2 x y + 3

$2 x y + 3$ and

0

$0$ .

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

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

Step 9.1.3.3

Subtract

2 x y

$2 x y$ from

2 x y

$2 x y$ .

\frac{d g}{d y} = 0 + 3

$\frac{d g}{d y} = 0 + 3$

Step 9.1.3.4

Add

0

$0$ and

3

$3$ .

\frac{d g}{d y} = 3

$\frac{d g}{d y} = 3$

\frac{d g}{d y} = 3

$\frac{d g}{d y} = 3$

\frac{d g}{d y} = 3

$\frac{d g}{d y} = 3$

\frac{d g}{d y} = 3

$\frac{d g}{d y} = 3$

Step 10

Find the antiderivative of

3

$3$ to find

g (y)

$g (y)$ .

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Step 10.1

Integrate both sides of

\frac{d g}{d y} = 3

$\frac{d g}{d y} = 3$ .

\int \frac{d g}{d y} d y = \int 3 d y

$\int \frac{d g}{d y} d y = \int 3 d y$

Step 10.2

Evaluate

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

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

g (y) = \int 3 d y

$g (y) = \int 3 d y$

Step 10.3

Apply the constant rule.

g (y) = 3 y + C

$g (y) = 3 y + C$

g (y) = 3 y + C

$g (y) = 3 y + C$

Step 11

Substitute for

g (y)

$g (y)$ in

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

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

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

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

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