Calculus Examples

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

Step 1

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

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Step 1.1

Rewrite.

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

Step 2

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

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Step 2.1

Differentiate $M$ with respect to $y$ .

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

Step 2.2

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 2.3

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 2.4

Multiply $2$ by $1$ .

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

Step 3

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

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Step 3.1

Differentiate $N$ with respect to $x$ .

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

Step 3.2

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

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

Step 3.3

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

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

Step 3.4

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

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

Step 3.5

Add $2 x$ and $0$ .

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

Step 4

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

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Step 4.1

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

$2 x = 2 x$

Step 4.2

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

$2 x = 2 x$ is an identity.

Step 5

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

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

Step 6

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

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Step 6.1

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

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

Step 6.2

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

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

Step 6.3

Simplify the answer.

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Step 6.3.1

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

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

Step 6.3.2

Simplify.

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Step 6.3.2.1

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

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

Step 6.3.2.2

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

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

Step 6.3.2.3

Move $2$ to the left of $y$ .

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

Step 6.3.2.4

Multiply $2$ by $y$ .

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

Step 6.3.2.5

Cancel the common factor of $2$ .

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Step 6.3.2.5.1

Cancel the common factor.

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

Step 6.3.2.5.2

Divide $y$ by $1$ .

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

Step 7

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

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

Step 8

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

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

Step 9

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

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Step 9.1

Differentiate $f$ with respect to $y$ .

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

Step 9.2

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

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

Step 9.3

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

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Step 9.3.1

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

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

Step 9.3.2

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

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

Step 9.3.3

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

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

Step 9.4

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

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

Step 9.5

Reorder terms.

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

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 $x^{2}$ from both sides of the equation.

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

Step 10.1.2

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

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Step 10.1.2.1

Subtract $x^{2}$ from $x^{2}$ .

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

Step 10.1.2.2

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

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

Step 11

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

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Step 11.1

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

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

Step 11.2

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

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

Step 11.3

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

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

Step 12

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

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

Step 13

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

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