Calculus Examples

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

Step 1

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

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Step 1.1

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

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

Step 1.2

Rewrite.

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

Step 2

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

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Step 2.1

Differentiate $M$ with respect to $y$ .

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

Step 2.2

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 2.3

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 2.4

Multiply $2$ by $- 1$ .

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

Step 3

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

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Step 3.1

Differentiate $N$ with respect to $x$ .

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

Step 3.2

Differentiate.

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Step 3.2.1

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

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

Step 3.2.2

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

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

Step 3.3

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

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Step 3.3.1

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

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

Step 3.3.2

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

$\frac{\partial N}{\partial x} = 2 x + 2 y \cdot 1$

Step 3.3.3

Multiply $2$ by $1$ .

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

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

$- 2 y = 2 x + 2 y$

Step 4.2

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

$- 2 y = 2 x + 2 y$ 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 $2 x + 2 y$ for $\frac{\partial N}{\partial x}$ .

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

Step 5.3

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

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Step 5.3.1

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

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

Step 5.3.2

Simplify the numerator.

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Step 5.3.2.1

Apply the distributive property.

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

Step 5.3.2.2

Multiply $2$ by $- 1$ .

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

Step 5.3.2.3

Multiply $2$ by $- 1$ .

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

Step 5.3.2.4

Subtract $2 y$ from $- 2 y$ .

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

Step 5.3.2.5

Factor $2$ out of $- 2 x - 4 y$ .

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Step 5.3.2.5.1

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

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

Step 5.3.2.5.2

Factor $2$ out of $- 4 y$ .

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

Step 5.3.2.5.3

Factor $2$ out of $2 (- x) + 2 (- 2 y)$ .

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

Step 5.3.3

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

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Step 5.3.3.1

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

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

Step 5.3.3.2

Factor $x$ out of $2 x y$ .

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

Step 5.3.3.3

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

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

Step 5.3.4

Cancel the common factor of $- x - 2 y$ and $x + 2 y$ .

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Step 5.3.4.1

Factor $- 1$ out of $- x$ .

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

Step 5.3.4.2

Factor $- 1$ out of $- 2 y$ .

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

Step 5.3.4.3

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

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

Step 5.3.4.4

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

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

Step 5.3.4.5

Cancel the common factor.

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

Step 5.3.4.6

Rewrite the expression.

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

Step 5.3.5

Multiply $2$ by $- 1$ .

$\frac{- 2}{x}$

Step 5.3.6

Move the negative in front of the fraction.

$- \frac{2}{x}$

Step 5.4

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

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

Step 6

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

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Step 6.1

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

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

Step 6.2

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

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

Step 6.3

Multiply $2$ by $- 1$ .

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

Step 6.4

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

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

Step 6.5

Simplify.

$μ (x, y) = e^{- 2 \ln (| x |)} + C$

Step 6.6

Simplify each term.

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Step 6.6.1

Simplify $- 2 \ln (| x |)$ by moving $- 2$ inside the logarithm.

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

Step 6.6.2

Exponentiation and log are inverse functions.

$μ (x, y) = {| x |}^{- 2} + C$

Step 6.6.3

Remove the absolute value in ${| x |}^{- 2}$ because exponentiations with even powers are always positive.

$μ (x, y) = x^{- 2} + C$

Step 6.6.4

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

$μ (x, y) = \frac{1}{x^{2}} + C$

Step 7

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

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Step 7.1

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

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

Step 7.2

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

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

Step 7.3

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

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

Step 7.4

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

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

Step 7.5

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

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Step 7.5.1

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

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

Step 7.5.2

Factor $x$ out of $2 x y$ .

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

Step 7.5.3

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

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

Step 7.6

Cancel the common factors.

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Step 7.6.1

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

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

Step 7.6.2

Cancel the common factor.

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

Step 7.6.3

Rewrite the expression.

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

Step 8

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

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

Step 9

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

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Step 9.1

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

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

Step 9.2

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

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

Step 9.3

Remove parentheses.

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

Step 9.4

Move $x^{2}$ out of the denominator by raising it to the $- 1$ power.

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

Step 9.5

Multiply the exponents in ${(x^{2})}^{- 1}$ .

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Step 9.5.1

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

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

Step 9.5.2

Multiply $2$ by $- 1$ .

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

Step 9.6

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

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

Step 9.7

Simplify the answer.

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Step 9.7.1

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

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

Step 9.7.2

Simplify.

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Step 9.7.2.1

Multiply $- 1$ by $- 1$ .

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

Step 9.7.2.2

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

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

Step 9.7.2.3

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

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

Step 10

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

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

Step 11

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

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

Step 12

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

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Step 12.1

Differentiate $f$ with respect to $y$ .

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

Step 12.2

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

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

Step 12.3

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

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Step 12.3.1

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

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

Step 12.3.2

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

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

Step 12.3.3

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

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

Step 12.3.4

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

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

Step 12.4

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

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

Step 12.5

Reorder terms.

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

Step 13

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

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Step 13.1

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

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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

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

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

Step 13.1.1.2

Combine the numerators over the common denominator.

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

Step 13.1.1.3

Simplify each term.

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Step 13.1.1.3.1

Apply the distributive property.

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

Step 13.1.1.3.2

Multiply $2$ by $- 1$ .

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

Step 13.1.1.4

Combine the opposite terms in $2 y - x - 2 y$ .

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Step 13.1.1.4.1

Subtract $2 y$ from $2 y$ .

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

Step 13.1.1.4.2

Add $- x$ and $0$ .

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

Step 13.1.1.5

Cancel the common factor of $x$ .

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Step 13.1.1.5.1

Cancel the common factor.

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

Step 13.1.1.5.2

Divide $- 1$ by $1$ .

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

Step 13.1.2

Add $1$ to both sides of the equation.

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

Step 14

Find the antiderivative of $1$ to find $g (y)$ .

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Step 14.1

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

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

Step 14.2

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

$g (y) = \int d y$

Step 14.3

Apply the constant rule.

$g (y) = y + C$

Step 15

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

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