Calculus Examples

$x d x + (x^{2} y + 4 y) d y = 0$

Step 1

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

Tap for more steps...

Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

Since $x$ is constant with respect to $y$ , the derivative of $x$ with respect to $y$ is $0$ .

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

Step 2

Find $\frac{\partial N}{\partial x}$ where $N (x, y) = x^{2} y + 4 y$ .

Tap for more steps...

Step 2.1

Differentiate $N$ with respect to $x$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [x^{2} y + 4 y]$

Step 2.2

By the Sum Rule, the derivative of $x^{2} y + 4 y$ with respect to $x$ is $\frac{d}{d x} [x^{2} y] + \frac{d}{d x} [4 y]$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [x^{2} y] + \frac{d}{d x} [4 y]$

Step 2.3

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

Tap for more steps...

Step 2.3.1

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

$\frac{\partial N}{\partial x} = y \frac{d}{d x} [x^{2}] + \frac{d}{d x} [4 y]$

Step 2.3.2

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

$\frac{\partial N}{\partial x} = y (2 x) + \frac{d}{d x} [4 y]$

Step 2.3.3

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

$\frac{\partial N}{\partial x} = 2 y x + \frac{d}{d x} [4 y]$

Step 2.4

Since $4 y$ is constant with respect to $x$ , the derivative of $4 y$ with respect to $x$ is $0$ .

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

Step 2.5

Simplify.

Tap for more steps...

Step 2.5.1

Add $2 y x$ and $0$ .

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

Step 2.5.2

Reorder the factors of $2 y x$ .

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

Step 3

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

Tap for more steps...

Step 3.1

Substitute $0$ for $\frac{\partial M}{\partial y}$ and $2 x y$ for $\frac{\partial N}{\partial x}$ .

$0 = 2 x y$

Step 3.2

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

$0 = 2 x y$ is not an identity.

Step 4

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

Tap for more steps...

Step 4.1

Substitute $0$ for $\frac{\partial M}{\partial y}$ .

$\frac{0 - \frac{\partial N}{\partial x}}{N}$

Step 4.2

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

$\frac{0 - (2 x y)}{N}$

Step 4.3

Substitute $x^{2} y + 4 y$ for $N$ .

Tap for more steps...

Step 4.3.1

Substitute $x^{2} y + 4 y$ for $N$ .

$\frac{0 - (2 x y)}{x^{2} y + 4 y}$

Step 4.3.2

Simplify the numerator.

Tap for more steps...

Step 4.3.2.1

Multiply $- 1$ by $2$ .

$\frac{0 - 2 x y}{x^{2} y + 4 y}$

Step 4.3.2.2

Subtract $2 x y$ from $0$ .

$\frac{- 2 x y}{x^{2} y + 4 y}$

Step 4.3.3

Factor $y$ out of $x^{2} y + 4 y$ .

Tap for more steps...

Step 4.3.3.1

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

$\frac{- 2 x y}{y x^{2} + 4 y}$

Step 4.3.3.2

Factor $y$ out of $4 y$ .

$\frac{- 2 x y}{y x^{2} + y \cdot 4}$

Step 4.3.3.3

Factor $y$ out of $y x^{2} + y \cdot 4$ .

$\frac{- 2 x y}{y (x^{2} + 4)}$

Step 4.3.4

Cancel the common factor of $y$ .

Tap for more steps...

Step 4.3.4.1

Cancel the common factor.

$\frac{- 2 x y}{y (x^{2} + 4)}$

Step 4.3.4.2

Rewrite the expression.

$\frac{- 2 x}{x^{2} + 4}$

Step 4.3.5

Move the negative in front of the fraction.

$- \frac{2 x}{x^{2} + 4}$

Step 4.4

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

$μ (x, y) = e^{\int - \frac{2 x}{x^{2} + 4} d x}$

Step 5

Evaluate the integral $e^{\int - \frac{2 x}{x^{2} + 4} d x}$ .

Tap for more steps...

Step 5.1

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

$μ (x, y) = e^{- \int \frac{2 x}{x^{2} + 4} d x}$

Step 5.2

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

$μ (x, y) = e^{- (2 \int \frac{x}{x^{2} + 4} d x)}$

Step 5.3

Multiply $2$ by $- 1$ .

$μ (x, y) = e^{- 2 \int \frac{x}{x^{2} + 4} d x}$

Step 5.4

Let $u = x^{2} + 4$ . Then $d u = 2 x d x$ , so $\frac{1}{2} d u = x d x$ . Rewrite using $u$ and $d$ $u$ .

Tap for more steps...

Step 5.4.1

Let $u = x^{2} + 4$ . Find $\frac{d u}{d x}$ .

Tap for more steps...

Step 5.4.1.1

Differentiate $x^{2} + 4$ .

$\frac{d}{d x} [x^{2} + 4]$

Step 5.4.1.2

By the Sum Rule, the derivative of $x^{2} + 4$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [4]$ .

$\frac{d}{d x} [x^{2}] + \frac{d}{d x} [4]$

Step 5.4.1.3

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

$2 x + \frac{d}{d x} [4]$

Step 5.4.1.4

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

$2 x + 0$

Step 5.4.1.5

Add $2 x$ and $0$ .

$2 x$

Step 5.4.2

Rewrite the problem using $u$ and $d u$ .

$μ (x, y) = e^{- 2 \int \frac{1}{u} \cdot \frac{1}{2} d u}$

Step 5.5

Simplify.

Tap for more steps...

Step 5.5.1

Multiply $\frac{1}{u}$ by $\frac{1}{2}$ .

$μ (x, y) = e^{- 2 \int \frac{1}{u \cdot 2} d u}$

Step 5.5.2

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

$μ (x, y) = e^{- 2 \int \frac{1}{2 u} d u}$

Step 5.6

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

$μ (x, y) = e^{- 2 (\frac{1}{2} \int \frac{1}{u} d u)}$

Step 5.7

Simplify.

Tap for more steps...

Step 5.7.1

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

$μ (x, y) = e^{\frac{- 2}{2} \int \frac{1}{u} d u}$

Step 5.7.2

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

Tap for more steps...

Step 5.7.2.1

Factor $2$ out of $- 2$ .

$μ (x, y) = e^{\frac{2 \cdot - 1}{2} \int \frac{1}{u} d u}$

Step 5.7.2.2

Cancel the common factors.

Tap for more steps...

Step 5.7.2.2.1

Factor $2$ out of $2$ .

$μ (x, y) = e^{\frac{2 \cdot - 1}{2 (1)} \int \frac{1}{u} d u}$

Step 5.7.2.2.2

Cancel the common factor.

$μ (x, y) = e^{\frac{2 \cdot - 1}{2 \cdot 1} \int \frac{1}{u} d u}$

Step 5.7.2.2.3

Rewrite the expression.

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

Step 5.7.2.2.4

Divide $- 1$ by $1$ .

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

Step 5.8

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

$μ (x, y) = e^{- (\ln (| u |) + C)}$

Step 5.9

Simplify.

$μ (x, y) = e^{- \ln (| u |)} + C$

Step 5.10

Replace all occurrences of $u$ with $x^{2} + 4$ .

$μ (x, y) = e^{- \ln (| x^{2} + 4 |)} + C$

Step 5.11

Simplify each term.

Tap for more steps...

Step 5.11.1

Simplify $- \ln (| x^{2} + 4 |)$ by moving $- 1$ inside the logarithm.

$μ (x, y) = e^{\ln ({| x^{2} + 4 |}^{- 1})} + C$

Step 5.11.2

Exponentiation and log are inverse functions.

$μ (x, y) = {| x^{2} + 4 |}^{- 1} + C$

Step 5.11.3

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

$μ (x, y) = \frac{1}{| x^{2} + 4 |} + C$

Step 6

Multiply both sides of $x d x + (x^{2} y + 4 y) d y = 0$ by the integration factor $\frac{1}{x^{2} + 4}$ .

Tap for more steps...

Step 6.1

Multiply $x$ by $\frac{1}{x^{2} + 4}$ .

$x \frac{1}{x^{2} + 4} d x + (x^{2} y + 4 y) d y = 0$

Step 6.2

Combine $x$ and $\frac{1}{x^{2} + 4}$ .

$\frac{x}{x^{2} + 4} d x + (x^{2} y + 4 y) d y = 0$

Step 6.3

Multiply $x^{2} y + 4 y$ by $\frac{1}{x^{2} + 4}$ .

$\frac{x}{x^{2} + 4} d x + (x^{2} y + 4 y) \frac{1}{x^{2} + 4} d y = 0$

Step 6.4

Multiply $x^{2} y + 4 y$ by $\frac{1}{x^{2} + 4}$ .

$\frac{x}{x^{2} + 4} d x + \frac{x^{2} y + 4 y}{x^{2} + 4} d y = 0$

Step 6.5

Factor $y$ out of $x^{2} y + 4 y$ .

Tap for more steps...

Step 6.5.1

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

$\frac{x}{x^{2} + 4} d x + \frac{y x^{2} + 4 y}{x^{2} + 4} d y = 0$

Step 6.5.2

Factor $y$ out of $4 y$ .

$\frac{x}{x^{2} + 4} d x + \frac{y x^{2} + y \cdot 4}{x^{2} + 4} d y = 0$

Step 6.5.3

Factor $y$ out of $y x^{2} + y \cdot 4$ .

$\frac{x}{x^{2} + 4} d x + \frac{y (x^{2} + 4)}{x^{2} + 4} d y = 0$

Step 6.6

Cancel the common factor of $x^{2} + 4$ .

Tap for more steps...

Step 6.6.1

Cancel the common factor.

$\frac{x}{x^{2} + 4} d x + \frac{y (x^{2} + 4)}{x^{2} + 4} d y = 0$

Step 6.6.2

Divide $y$ by $1$ .

$\frac{x}{x^{2} + 4} d x + y d y = 0$

Step 7

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

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

Step 8

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

Tap for more steps...

Step 8.1

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

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

Step 9

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

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

Step 10

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

$\frac{\partial f}{\partial x} = \frac{x}{x^{2} + 4}$

Step 11

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

Tap for more steps...

Step 11.1

Differentiate $f$ with respect to $x$ .

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

Step 11.2

By the Sum Rule, the derivative of $\frac{1}{2} y^{2} + g (x)$ with respect to $x$ is $\frac{d}{d x} [\frac{1}{2} y^{2}] + \frac{d}{d x} [g (x)]$ .

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

Step 11.3

Since $\frac{1}{2} y^{2}$ is constant with respect to $x$ , the derivative of $\frac{1}{2} y^{2}$ with respect to $x$ is $0$ .

$0 + \frac{d}{d x} [g (x)] = \frac{x}{x^{2} + 4}$

Step 11.4

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

$0 + \frac{d g}{d x} = \frac{x}{x^{2} + 4}$

Step 11.5

Add $0$ and $\frac{d g}{d x}$ .

$\frac{d g}{d x} = \frac{x}{x^{2} + 4}$

Step 12

Find the antiderivative of $\frac{x}{x^{2} + 4}$ to find $g (x)$ .

Tap for more steps...

Step 12.1

Integrate both sides of $\frac{d g}{d x} = \frac{x}{x^{2} + 4}$ .

$\int \frac{d g}{d x} d x = \int \frac{x}{x^{2} + 4} d x$

Step 12.2

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

$g (x) = \int \frac{x}{x^{2} + 4} d x$

Step 12.3

Let $u = x^{2} + 4$ . Then $d u = 2 x d x$ , so $\frac{1}{2} d u = x d x$ . Rewrite using $u$ and $d$ $u$ .

Tap for more steps...

Step 12.3.1

Let $u = x^{2} + 4$ . Find $\frac{d u}{d x}$ .

Tap for more steps...

Step 12.3.1.1

Differentiate $x^{2} + 4$ .

$\frac{d}{d x} [x^{2} + 4]$

Step 12.3.1.2

By the Sum Rule, the derivative of $x^{2} + 4$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [4]$ .

$\frac{d}{d x} [x^{2}] + \frac{d}{d x} [4]$

Step 12.3.1.3

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

$2 x + \frac{d}{d x} [4]$

Step 12.3.1.4

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

$2 x + 0$

Step 12.3.1.5

Add $2 x$ and $0$ .

$2 x$

Step 12.3.2

Rewrite the problem using $u$ and $d u$ .

$g (x) = \int \frac{1}{u} \cdot \frac{1}{2} d u$

Step 12.4

Simplify.

Tap for more steps...

Step 12.4.1

Multiply $\frac{1}{u}$ by $\frac{1}{2}$ .

$g (x) = \int \frac{1}{u \cdot 2} d u$

Step 12.4.2

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

$g (x) = \int \frac{1}{2 \cdot u} d u$

Step 12.4.3

Multiply $2$ by $u$ .

$g (x) = \int \frac{1}{2 u} d u$

Step 12.5

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

$g (x) = \frac{1}{2} \int \frac{1}{u} d u$

Step 12.6

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

$g (x) = \frac{1}{2} (\ln (| u |) + C)$

Step 12.7

Simplify.

$g (x) = \frac{1}{2} \ln (| u |) + C$

Step 12.8

Replace all occurrences of $u$ with $x^{2} + 4$ .

$g (x) = \frac{1}{2} \ln (| x^{2} + 4 |) + C$

Step 13

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

$f (x, y) = \frac{1}{2} y^{2} + \frac{1}{2} \ln (| x^{2} + 4 |) + C$

Step 14

Simplify $f (x, y) = \frac{1}{2} y^{2} + \frac{1}{2} \ln (| x^{2} + 4 |) + C$ .

Tap for more steps...

Step 14.1

Simplify each term.

Tap for more steps...

Step 14.1.1

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

$f (x, y) = \frac{y^{2}}{2} + \frac{1}{2} \ln (| x^{2} + 4 |) + C$

Step 14.1.2

Simplify $\frac{1}{2} \ln (| x^{2} + 4 |)$ by moving $\frac{1}{2}$ inside the logarithm.

$f (x, y) = \frac{y^{2}}{2} + \ln ({| x^{2} + 4 |}^{\frac{1}{2}}) + C$

Step 14.2

To write $\ln ({| x^{2} + 4 |}^{\frac{1}{2}})$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f (x, y) = \frac{y^{2}}{2} + \ln ({| x^{2} + 4 |}^{\frac{1}{2}}) \cdot \frac{2}{2} + C$

Step 14.3

Combine $\ln ({| x^{2} + 4 |}^{\frac{1}{2}})$ and $\frac{2}{2}$ .

$f (x, y) = \frac{y^{2}}{2} + \frac{\ln ({| x^{2} + 4 |}^{\frac{1}{2}}) \cdot 2}{2} + C$

Step 14.4

Combine the numerators over the common denominator.

$f (x, y) = \frac{y^{2} + \ln ({| x^{2} + 4 |}^{\frac{1}{2}}) \cdot 2}{2} + C$

Step 14.5

Simplify the numerator.

Tap for more steps...

Step 14.5.1

Multiply $\ln ({| x^{2} + 4 |}^{\frac{1}{2}}) \cdot 2$ .

Tap for more steps...

Step 14.5.1.1

Reorder $\ln ({| x^{2} + 4 |}^{\frac{1}{2}})$ and $2$ .

$f (x, y) = \frac{y^{2} + 2 \cdot \ln ({| x^{2} + 4 |}^{\frac{1}{2}})}{2} + C$

Step 14.5.1.2

Simplify $2 \cdot \ln ({| x^{2} + 4 |}^{\frac{1}{2}})$ by moving $2$ inside the logarithm.

$f (x, y) = \frac{y^{2} + \ln ({({| x^{2} + 4 |}^{\frac{1}{2}})}^{2})}{2} + C$

Step 14.5.2

Multiply the exponents in ${({| x^{2} + 4 |}^{\frac{1}{2}})}^{2}$ .

Tap for more steps...

Step 14.5.2.1

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

$f (x, y) = \frac{y^{2} + \ln ({| x^{2} + 4 |}^{\frac{1}{2} \cdot 2})}{2} + C$

Step 14.5.2.2

Cancel the common factor of $2$ .

Tap for more steps...

Step 14.5.2.2.1

Cancel the common factor.

$f (x, y) = \frac{y^{2} + \ln ({| x^{2} + 4 |}^{\frac{1}{2} \cdot 2})}{2} + C$

Step 14.5.2.2.2

Rewrite the expression.

$f (x, y) = \frac{y^{2} + \ln ({| x^{2} + 4 |}^{1})}{2} + C$

Step 14.5.3

Simplify.

$f (x, y) = \frac{y^{2} + \ln (| x^{2} + 4 |)}{2} + C$