Calculus Examples

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

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.

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

Step 1.2

Rewrite.

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

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

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 2.7

Multiply $2$ by $- 1$ .

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

Step 3

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

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Step 3.1

Differentiate $N$ with respect to $x$ .

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

Step 3.2

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

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

Step 3.3

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

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

Step 3.4

Multiply $y$ by $1$ .

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

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

$- 2 y = y$

Step 4.2

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

$- 2 y = y$ is not an identity.

Step 5

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

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Step 5.1

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

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

Step 5.2

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

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

Step 5.3

Substitute $x y$ for $N$ .

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Step 5.3.1

Substitute $x y$ for $N$ .

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

Step 5.3.2

Subtract $y$ from $- 2 y$ .

$\frac{- 3 y}{x y}$

Step 5.3.3

Cancel the common factor of $y$ .

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Step 5.3.3.1

Cancel the common factor.

$\frac{- 3 y}{x y}$

Step 5.3.3.2

Rewrite the expression.

$\frac{- 3}{x}$

Step 5.3.4

Move the negative in front of the fraction.

$- \frac{3}{x}$

Step 5.4

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

$μ (x, y) = e^{\int - \frac{3}{x} d x}$

Step 6

Evaluate the integral $e^{\int - \frac{3}{x} d x}$ .

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Step 6.1

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

$μ (x, y) = e^{- \int \frac{3}{x} d x}$

Step 6.2

Since $3$ is constant with respect to $x$ , move $3$ out of the integral.

$μ (x, y) = e^{- (3 \int \frac{1}{x} d x)}$

Step 6.3

Multiply $3$ by $- 1$ .

$μ (x, y) = e^{- 3 \int \frac{1}{x} d x}$

Step 6.4

The integral of $\frac{1}{x}$ with respect to $x$ is $\ln (| x |)$ .

$μ (x, y) = e^{- 3 (\ln (| x |) + C)}$

Step 6.5

Simplify.

$μ (x, y) = e^{- 3 \ln (| x |)} + C$

Step 6.6

Simplify each term.

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Step 6.6.1

Simplify $- 3 \ln (| x |)$ by moving $- 3$ inside the logarithm.

$μ (x, y) = e^{\ln ({| x |}^{- 3})} + C$

Step 6.6.2

Exponentiation and log are inverse functions.

$μ (x, y) = {| x |}^{- 3} + C$

Step 6.6.3

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 7

Multiply both sides of $- (x^{2} + y^{2}) d x + x y d y = 0$ by the integration factor $\frac{1}{x^{3}}$ .

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Step 7.1

Multiply $- (x^{2} + y^{2})$ by $\frac{1}{x^{3}}$ .

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

Step 7.2

Apply the distributive property.

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

Step 7.3

Multiply $- x^{2} - y^{2}$ by $\frac{1}{x^{3}}$ .

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

Step 7.4

Factor $- 1$ out of $- x^{2}$ .

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

Step 7.5

Factor $- 1$ out of $- y^{2}$ .

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

Step 7.6

Factor $- 1$ out of $- (x^{2}) - (y^{2})$ .

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

Step 7.7

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

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

Step 7.8

Move the negative in front of the fraction.

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

Step 7.9

Multiply $x y$ by $\frac{1}{x^{3}}$ .

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

Step 7.10

Cancel the common factor of $x$ .

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Step 7.10.1

Factor $x$ out of $x y$ .

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

Step 7.10.2

Factor $x$ out of $x^{3}$ .

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

Step 7.10.3

Cancel the common factor.

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

Step 7.10.4

Rewrite the expression.

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

Step 7.11

Combine $y$ and $\frac{1}{x^{2}}$ .

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

Step 8

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

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

Step 9

Integrate $N (x, y) = \frac{y}{x^{2}}$ to find $f (x, y)$ .

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Step 9.1

Since $\frac{1}{x^{2}}$ is constant with respect to $y$ , move $\frac{1}{x^{2}}$ out of the integral.

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

Step 9.2

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

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

Step 9.3

Simplify the answer.

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Step 9.3.1

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

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

Step 9.3.2

Simplify.

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Step 9.3.2.1

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

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

Step 9.3.2.2

Move $2$ to the left of $x^{2}$ .

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

Step 9.3.2.3

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

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

Step 9.3.2.4

Combine $\frac{1}{2 x^{2}}$ and $y^{2}$ .

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

Step 10

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

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

Step 11

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

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

Step 12

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

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Step 12.1

Differentiate $f$ with respect to $x$ .

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

Step 12.2

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

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

Step 12.3

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

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Step 12.3.1

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

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

Step 12.3.2

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

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

Step 12.3.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- 1}$ and $g (x) = x^{2}$ .

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Step 12.3.3.1

To apply the Chain Rule, set $u$ as $x^{2}$ .

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

Step 12.3.3.2

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

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

Step 12.3.3.3

Replace all occurrences of $u$ with $x^{2}$ .

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

Step 12.3.4

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

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

Step 12.3.5

Multiply the exponents in ${(x^{2})}^{- 2}$ .

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Step 12.3.5.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 12.3.5.2

Multiply $2$ by $- 2$ .

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

Step 12.3.6

Multiply $2$ by $- 1$ .

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

Step 12.3.7

Raise $x$ to the power of $1$ .

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

Step 12.3.8

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 12.3.9

Subtract $4$ from $1$ .

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

Step 12.3.10

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

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

Step 12.3.11

Combine $\frac{- 2 y^{2}}{2}$ and $x^{- 3}$ .

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

Step 12.3.12

Move $x^{- 3}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 12.3.13

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

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Step 12.3.13.1

Factor $2$ out of $- 2 y^{2}$ .

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

Step 12.3.13.2

Cancel the common factors.

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Step 12.3.13.2.1

Factor $2$ out of $2 x^{3}$ .

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

Step 12.3.13.2.2

Cancel the common factor.

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

Step 12.3.13.2.3

Rewrite the expression.

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

Step 12.3.14

Move the negative in front of the fraction.

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

Step 12.4

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

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

Step 12.5

Reorder terms.

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

Step 13

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

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Step 13.1

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

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Step 13.1.1

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

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Step 13.1.1.1

Add $\frac{x^{2} + y^{2}}{x^{3}}$ to both sides of the equation.

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

Step 13.1.1.2

Combine the numerators over the common denominator.

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

Step 13.1.1.3

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

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Step 13.1.1.3.1

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

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

Step 13.1.1.3.2

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

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

Step 13.1.1.4

Cancel the common factor of $x^{2}$ and $x^{3}$ .

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Step 13.1.1.4.1

Multiply by $1$ .

$\frac{d g}{d x} + \frac{x^{2} \cdot 1}{x^{3}} = 0$

Step 13.1.1.4.2

Cancel the common factors.

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Step 13.1.1.4.2.1

Factor $x^{2}$ out of $x^{3}$ .

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

Step 13.1.1.4.2.2

Cancel the common factor.

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

Step 13.1.1.4.2.3

Rewrite the expression.

$\frac{d g}{d x} + \frac{1}{x} = 0$

Step 13.1.2

Subtract $\frac{1}{x}$ from both sides of the equation.

$\frac{d g}{d x} = - \frac{1}{x}$

Step 14

Find the antiderivative of $- \frac{1}{x}$ to find $g (x)$ .

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Step 14.1

Integrate both sides of $\frac{d g}{d x} = - \frac{1}{x}$ .

$\int \frac{d g}{d x} d x = \int - \frac{1}{x} d x$

Step 14.2

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

$g (x) = \int - \frac{1}{x} d x$

Step 14.3

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

$g (x) = - \int \frac{1}{x} d x$

Step 14.4

The integral of $\frac{1}{x}$ with respect to $x$ is $\ln (| x |)$ .

$g (x) = - (\ln (| x |) + C)$

Step 14.5

Simplify.

$g (x) = - \ln (| x |) + C$

Step 15

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

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