Calculus Examples

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

Step 1

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

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Step 1.1

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

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

Step 2

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

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Step 2.1

Differentiate $M$ with respect to $y$ .

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

Step 2.2

Differentiate.

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Step 2.2.1

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

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

Step 2.2.2

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

$\frac{\partial M}{\partial y} = 0 + \frac{d}{d y} [- y^{2}]$

Step 2.3

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

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Step 2.3.1

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

$\frac{\partial M}{\partial y} = 0 - \frac{d}{d y} [y^{2}]$

Step 2.3.2

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

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

Step 2.3.3

Multiply $2$ by $- 1$ .

$\frac{\partial M}{\partial y} = 0 - 2 y$

Step 2.4

Subtract $2 y$ from $0$ .

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

Step 3

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

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Step 3.1

Differentiate $N$ with respect to $x$ .

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

Step 3.2

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

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

Step 3.3

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

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

Step 3.4

Multiply $- 2$ by $1$ .

$\frac{\partial N}{\partial x} = - 2 y$

Step 4

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

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Step 4.1

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

$- 2 y = - 2 y$

Step 4.2

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

$- 2 y = - 2 y$ is an identity.

Step 5

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

$f (x, y) = \int - 2 x y d y$

Step 6

Integrate $N (x, y) = - 2 x y$ to find $f (x, y)$ .

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Step 6.1

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

$f (x, y) = - 2 x \int y d y$

Step 6.2

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

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

Step 6.3

Simplify the answer.

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Step 6.3.1

Rewrite $- 2 x (\frac{1}{2} y^{2} + C)$ as $- 2 x \frac{1}{2} y^{2} + C$ .

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

Step 6.3.2

Simplify.

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Step 6.3.2.1

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

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

Step 6.3.2.2

Combine $x$ and $\frac{- 2}{2}$ .

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

Step 6.3.2.3

Move $- 2$ to the left of $x$ .

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

Step 6.3.2.4

Cancel the common factor of $- 2$ and $2$ .

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Step 6.3.2.4.1

Factor $2$ out of $- 2 x$ .

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

Step 6.3.2.4.2

Cancel the common factors.

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Step 6.3.2.4.2.1

Factor $2$ out of $2$ .

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

Step 6.3.2.4.2.2

Cancel the common factor.

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

Step 6.3.2.4.2.3

Rewrite the expression.

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

Step 6.3.2.4.2.4

Divide $- x$ by $1$ .

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

Step 7

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

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

Step 8

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

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

Step 9

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

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Step 9.1

Differentiate $f$ with respect to $x$ .

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

Step 9.2

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

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

Step 9.3

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

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Step 9.3.1

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

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

Step 9.3.2

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

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

Step 9.3.3

Multiply $- 1$ by $1$ .

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

Step 9.4

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

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

Step 9.5

Reorder terms.

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

Step 10

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

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Step 10.1

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

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Step 10.1.1

Add $y^{2}$ to both sides of the equation.

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

Step 10.1.2

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

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Step 10.1.2.1

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

$\frac{d g}{d x} = x^{2} + 0$

Step 10.1.2.2

Add $x^{2}$ and $0$ .

$\frac{d g}{d x} = x^{2}$

Step 11

Find the antiderivative of $x^{2}$ to find $g (x)$ .

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Step 11.1

Integrate both sides of $\frac{d g}{d x} = x^{2}$ .

$\int \frac{d g}{d x} d x = \int x^{2} d x$

Step 11.2

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

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

Step 11.3

By the Power Rule, the integral of $x^{2}$ with respect to $x$ is $\frac{1}{3} x^{3}$ .

$g (x) = \frac{1}{3} x^{3} + C$

Step 12

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

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

Step 13

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

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