Calculus Examples

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

Step 1

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

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Step 1.1

Subtract $(x^{2} - y^{2}) d x$ from both sides of the equation.

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

Step 1.2

Rewrite.

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

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

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

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

Step 2.3

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 2.4

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 2.5

Add $0$ and $\frac{d}{d y} [- y^{2}]$ .

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

Step 2.6

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

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

Step 2.7

Multiply.

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Step 2.7.1

Multiply $- 1$ by $- 1$ .

$\frac{\partial M}{\partial y} = 1 \frac{d}{d y} [y^{2}]$

Step 2.7.2

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

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

Step 2.8

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 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 $M (x, y)$ .

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

Step 6

Integrate $M (x, y) = - (x^{2} - y^{2})$ to find $f (x, y)$ .

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Step 6.1

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

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

Step 6.2

Split the single integral into multiple integrals.

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

Step 6.3

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

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

Step 6.4

Apply the constant rule.

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

Step 6.5

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

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

Step 6.6

Simplify.

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

Step 7

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

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

Step 8

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

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

Step 9

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

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Step 9.1

Differentiate $f$ with respect to $y$ .

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

Step 9.2

Differentiate using the Sum Rule.

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Step 9.2.1

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

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

Step 9.2.2

By the Sum Rule, the derivative of $- (\frac{x^{3}}{3} - y^{2} x) + g (y)$ with respect to $y$ is $\frac{d}{d y} [- (\frac{x^{3}}{3} - y^{2} x)] + \frac{d}{d y} [g (y)]$ .

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

Step 9.3

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

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Step 9.3.1

Since $- 1$ is constant with respect to $y$ , the derivative of $- (\frac{x^{3}}{3} - y^{2} x)$ with respect to $y$ is $- \frac{d}{d y} [\frac{x^{3}}{3} - y^{2} x]$ .

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

Step 9.3.2

By the Sum Rule, the derivative of $\frac{x^{3}}{3} - y^{2} x$ with respect to $y$ is $\frac{d}{d y} [\frac{x^{3}}{3}] + \frac{d}{d y} [- y^{2} x]$ .

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

Step 9.3.3

Since $\frac{x^{3}}{3}$ is constant with respect to $y$ , the derivative of $\frac{x^{3}}{3}$ with respect to $y$ is $0$ .

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

Step 9.3.4

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

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

Step 9.3.5

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

$- (0 - x (2 y)) + \frac{d}{d y} [g (y)] = 2 x y$

Step 9.3.6

Multiply $2$ by $- 1$ .

$- (0 - 2 x y) + \frac{d}{d y} [g (y)] = 2 x y$

Step 9.3.7

Subtract $2 x y$ from $0$ .

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

Step 9.3.8

Multiply $- 2$ by $- 1$ .

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

Step 9.4

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

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

Step 9.5

Reorder terms.

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

Step 10

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

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Step 10.1

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

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Step 10.1.1

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

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

Step 10.1.2

Subtract $2 x y$ from $2 x y$ .

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

Step 11

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

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Step 11.1

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

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

Step 11.2

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

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

Step 11.3

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

$g (y) = 0 + C$

Step 11.4

Add $0$ and $C$ .

$g (y) = C$

Step 12

Substitute for $g (y)$ in $f (x, y) = - (\frac{1}{3} x^{3} - y^{2} x) + g (y)$ .

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

Step 13

Simplify each term.

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Step 13.1

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

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

Step 13.2

Apply the distributive property.

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

Step 13.3

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

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Step 13.3.1

Multiply $- 1$ by $- 1$ .

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

Step 13.3.2

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

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