Calculus Examples

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

Step 1

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

Tap for more steps...

Step 1.1

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

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

Step 1.2

Rewrite.

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

Step 2

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

Tap for more steps...

Step 2.1

Differentiate $M$ with respect to $y$ .

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

Step 2.2

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

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

Step 2.3

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

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

Step 2.4

Multiply $- 1$ by $1$ .

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

Step 3

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

Tap for more steps...

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

Tap for more steps...

Step 4.1

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

$- x = 2 x$

Step 4.2

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

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

Step 5

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

Tap for more steps...

Step 5.1

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

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

Step 5.2

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

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

Step 5.3

Substitute $- x y$ for $M$ .

Tap for more steps...

Step 5.3.1

Substitute $- x y$ for $M$ .

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

Step 5.3.2

Simplify the numerator.

Tap for more steps...

Step 5.3.2.1

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

Tap for more steps...

Step 5.3.2.1.1

Factor $x$ out of $2 x$ .

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

Step 5.3.2.1.2

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

$\frac{x \cdot 2 + x (- - 1)}{- x y}$

Step 5.3.2.1.3

Factor $x$ out of $x \cdot 2 + x (- - 1)$ .

$\frac{x (2 - - 1)}{- x y}$

Step 5.3.2.2

Multiply $- 1$ by $- 1$ .

$\frac{x (2 + 1)}{- x y}$

Step 5.3.2.3

Add $2$ and $1$ .

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

Step 5.3.3

Cancel the common factor of $x$ .

Tap for more steps...

Step 5.3.3.1

Cancel the common factor.

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

Step 5.3.3.2

Rewrite the expression.

$\frac{3}{- 1 y}$

Step 5.3.4

Substitute $- x y$ for $M$ .

$- \frac{3}{y}$

Step 5.4

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

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

Step 6

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

Tap for more steps...

Step 6.1

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

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

Step 6.2

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

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

Step 6.3

Multiply $3$ by $- 1$ .

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

Step 6.4

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

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

Step 6.5

Simplify.

$μ (x, y) = e^{- 3 \ln (| y |)} + C$

Step 6.6

Simplify each term.

Tap for more steps...

Step 6.6.1

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

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

Step 6.6.2

Exponentiation and log are inverse functions.

$μ (x, y) = {| y |}^{- 3} + C$

Step 6.6.3

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

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

Step 7

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

Tap for more steps...

Step 7.1

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

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

Step 7.2

Cancel the common factor of $y$ .

Tap for more steps...

Step 7.2.1

Factor $y$ out of $- x y$ .

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

Step 7.2.2

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

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

Step 7.2.3

Cancel the common factor.

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

Step 7.2.4

Rewrite the expression.

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

Step 7.3

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

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

Step 7.4

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

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

Step 7.5

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

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

Step 8

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

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

Step 9

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

Tap for more steps...

Step 9.1

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

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

Step 9.2

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

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

Step 9.3

Remove parentheses.

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

Step 9.4

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

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

Step 9.5

Simplify the answer.

Tap for more steps...

Step 9.5.1

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

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

Step 9.5.2

Simplify.

Tap for more steps...

Step 9.5.2.1

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

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

Step 9.5.2.2

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

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

Step 10

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

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

Step 11

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

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

Step 12

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

Tap for more steps...

Step 12.1

Differentiate $f$ with respect to $y$ .

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

Step 12.2

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

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

Step 12.3

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

Tap for more steps...

Step 12.3.1

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

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

Step 12.3.2

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

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

Step 12.3.3

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

Tap for more steps...

Step 12.3.3.1

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

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

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 $n = - 1$ .

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

Step 12.3.3.3

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

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

Step 12.3.4

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

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

Step 12.3.5

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

Tap for more steps...

Step 12.3.5.1

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

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

Step 12.3.5.2

Multiply $2$ by $- 2$ .

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

Step 12.3.6

Multiply $2$ by $- 1$ .

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

Step 12.3.7

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

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

Step 12.3.8

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

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

Step 12.3.9

Subtract $4$ from $1$ .

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

Step 12.3.10

Multiply $- 2$ by $- 1$ .

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

Step 12.3.11

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

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

Step 12.3.12

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

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

Step 12.3.13

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

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

Step 12.3.14

Cancel the common factor of $2$ .

Tap for more steps...

Step 12.3.14.1

Cancel the common factor.

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

Step 12.3.14.2

Rewrite the expression.

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

Step 12.4

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

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

Step 12.5

Reorder terms.

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

Step 13

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

Tap for more steps...

Step 13.1

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

Tap for more steps...

Step 13.1.1

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

Tap for more steps...

Step 13.1.1.1

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

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

Step 13.1.1.2

Combine the numerators over the common denominator.

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

Step 13.1.1.3

Apply the distributive property.

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

Step 13.1.1.4

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

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

Step 13.1.1.5

Subtract $y^{2}$ from $0$ .

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

Step 13.1.1.6

Simplify each term.

Tap for more steps...

Step 13.1.1.6.1

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

Tap for more steps...

Step 13.1.1.6.1.1

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

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

Step 13.1.1.6.1.2

Cancel the common factors.

Tap for more steps...

Step 13.1.1.6.1.2.1

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

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

Step 13.1.1.6.1.2.2

Cancel the common factor.

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

Step 13.1.1.6.1.2.3

Rewrite the expression.

$\frac{d g}{d y} + \frac{- 1}{y} = 0$

Step 13.1.1.6.2

Move the negative in front of the fraction.

$\frac{d g}{d y} - \frac{1}{y} = 0$

Step 13.1.2

Add $\frac{1}{y}$ to both sides of the equation.

$\frac{d g}{d y} = \frac{1}{y}$

Step 14

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

Tap for more steps...

Step 14.1

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

$\int \frac{d g}{d y} d y = \int \frac{1}{y} d y$

Step 14.2

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

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

Step 14.3

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

$g (y) = \ln (| y |) + C$

Step 15

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

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