Calculus Examples

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

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

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

$\frac{\partial M}{\partial y} = 1$

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

Differentiate.

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Step 2.2.1

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

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

Step 2.2.2

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

$\frac{\partial N}{\partial x} = 1 + \frac{d}{d x} [2 x y^{2}] + \frac{d}{d x} [- 2 y]$

Step 2.3

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

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Step 2.3.1

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

$\frac{\partial N}{\partial x} = 1 + 2 y^{2} \frac{d}{d x} [x] + \frac{d}{d x} [- 2 y]$

Step 2.3.2

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

$\frac{\partial N}{\partial x} = 1 + 2 y^{2} \cdot 1 + \frac{d}{d x} [- 2 y]$

Step 2.3.3

Multiply $2$ by $1$ .

$\frac{\partial N}{\partial x} = 1 + 2 y^{2} + \frac{d}{d x} [- 2 y]$

Step 2.4

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

$\frac{\partial N}{\partial x} = 1 + 2 y^{2} + 0$

Step 2.5

Simplify.

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Step 2.5.1

Add $1 + 2 y^{2}$ and $0$ .

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

Step 2.5.2

Reorder terms.

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

Step 3

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

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Step 3.1

Substitute $1$ for $\frac{\partial M}{\partial y}$ and $2 y^{2} + 1$ for $\frac{\partial N}{\partial x}$ .

$1 = 2 y^{2} + 1$

Step 3.2

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

$1 = 2 y^{2} + 1$ is not an identity.

Step 4

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

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Step 4.1

Substitute $2 y^{2} + 1$ for $\frac{\partial N}{\partial x}$ .

$\frac{2 y^{2} + 1 - \frac{\partial M}{\partial y}}{M}$

Step 4.2

Substitute $1$ for $\frac{\partial M}{\partial y}$ .

$\frac{2 y^{2} + 1 - 1}{M}$

Step 4.3

Substitute $y$ for $M$ .

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Step 4.3.1

Substitute $y$ for $M$ .

$\frac{2 y^{2} + 1 - 1}{y}$

Step 4.3.2

Simplify the numerator.

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Step 4.3.2.1

Subtract $1$ from $1$ .

$\frac{2 y^{2} + 0}{y}$

Step 4.3.2.2

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

$\frac{2 y^{2}}{y}$

Step 4.3.3

Substitute $y$ for $M$ .

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Step 4.3.3.1

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

$\frac{y (2 y)}{y}$

Step 4.3.3.2

Cancel the common factors.

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Step 4.3.3.2.1

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

$\frac{y (2 y)}{y^{1}}$

Step 4.3.3.2.2

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

$\frac{y (2 y)}{y \cdot 1}$

Step 4.3.3.2.3

Cancel the common factor.

$\frac{y (2 y)}{y \cdot 1}$

Step 4.3.3.2.4

Rewrite the expression.

$\frac{2 y}{1}$

Step 4.3.3.2.5

Divide $2 y$ by $1$ .

$2 y$

Step 4.4

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

$μ (x, y) = e^{\int 2 y d y}$

Step 5

Evaluate the integral $e^{\int 2 y d y}$ .

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Step 5.1

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

$μ (x, y) = e^{2 \int y d y}$

Step 5.2

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

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

Step 5.3

Simplify the answer.

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Step 5.3.1

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

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

Step 5.3.2

Simplify.

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Step 5.3.2.1

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

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

Step 5.3.2.2

Cancel the common factor of $2$ .

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Step 5.3.2.2.1

Cancel the common factor.

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

Step 5.3.2.2.2

Rewrite the expression.

$μ (x, y) = e^{1 y^{2}} + C$

Step 5.3.2.3

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

$μ (x, y) = e^{y^{2}} + C$

Step 6

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

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Step 6.1

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

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

Step 6.2

Multiply $x + 2 x y^{2} - 2 y$ by $e^{y^{2}}$ .

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

Step 6.3

Apply the distributive property.

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

Step 7

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

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

Step 8

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

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Step 8.1

Apply the constant rule.

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

Step 9

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

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

Step 10

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

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

Step 11

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

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Step 11.1

Differentiate $f$ with respect to $y$ .

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

Step 11.2

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

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

Step 11.3

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

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Step 11.3.1

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

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

Step 11.3.2

Differentiate using the Product Rule which states that $\frac{d}{d y} [f (y) g (y)]$ is $f (y) \frac{d}{d y} [g (y)] + g (y) \frac{d}{d y} [f (y)]$ where $f (y) = y$ and $g (y) = e^{y^{2}}$ .

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

Step 11.3.3

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

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Step 11.3.3.1

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

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

Step 11.3.3.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

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

Step 11.3.3.3

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

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

Step 11.3.4

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

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

Step 11.3.5

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

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

Step 11.3.6

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

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

Step 11.3.7

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

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

Step 11.3.8

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

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

Step 11.3.9

Add $1$ and $1$ .

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

Step 11.3.10

Move $2$ to the left of $e^{y^{2}}$ .

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

Step 11.3.11

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

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

Step 11.4

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

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

Step 11.5

Simplify.

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Step 11.5.1

Apply the distributive property.

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

Step 11.5.2

Reorder terms.

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

Step 11.5.3

Reorder factors in $\frac{d g}{d y} + 2 e^{y^{2}} y^{2} x + e^{y^{2}} x$ .

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

Step 12

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

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Step 12.1

Move all terms not containing $\frac{d g}{d y}$ to the right side of the equation.

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Step 12.1.1

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

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

Step 12.1.2

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

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

Step 12.1.3

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

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Step 12.1.3.1

Reorder the factors in the terms $2 x y^{2} e^{y^{2}}$ and $- 2 y^{2} x e^{y^{2}}$ .

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

Step 12.1.3.2

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

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

Step 12.1.3.3

Add $x e^{y^{2}} - 2 y e^{y^{2}}$ and $0$ .

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

Step 12.1.3.4

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

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

Step 12.1.3.5

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

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

Step 13

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

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Step 13.1

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

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

Step 13.2

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

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

Step 13.3

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

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

Step 13.4

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

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Step 13.4.1

Let $u = y^{2}$ . Find $\frac{d u}{d y}$ .

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Step 13.4.1.1

Differentiate $y^{2}$ .

$\frac{d}{d y} [y^{2}]$

Step 13.4.1.2

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

$2 y$

Step 13.4.2

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

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

Step 13.5

Combine $e^{u}$ and $\frac{1}{2}$ .

$g (y) = - 2 \int \frac{e^{u}}{2} d u$

Step 13.6

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

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

Step 13.7

Simplify.

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Step 13.7.1

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

$g (y) = \frac{- 2}{2} \int e^{u} d u$

Step 13.7.2

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

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Step 13.7.2.1

Factor $2$ out of $- 2$ .

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

Step 13.7.2.2

Cancel the common factors.

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Step 13.7.2.2.1

Factor $2$ out of $2$ .

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

Step 13.7.2.2.2

Cancel the common factor.

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

Step 13.7.2.2.3

Rewrite the expression.

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

Step 13.7.2.2.4

Divide $- 1$ by $1$ .

$g (y) = - \int e^{u} d u$

Step 13.8

The integral of $e^{u}$ with respect to $u$ is $e^{u}$ .

$g (y) = - (e^{u} + C)$

Step 13.9

Simplify.

$g (y) = - e^{u} + C$

Step 13.10

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

$g (y) = - e^{y^{2}} + C$

Step 14

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

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

Step 15

Reorder factors in $f (x, y) = y e^{y^{2}} x - e^{y^{2}} + C$ .

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