Calculus Examples

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

Step 1

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

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Step 1.1

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

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

Step 2

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

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Step 2.1

Differentiate $M$ with respect to $y$ .

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

Step 2.2

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 3

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

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Step 3.1

Differentiate $N$ with respect to $x$ .

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

Step 3.2

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

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

Step 3.3

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

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

Step 3.4

Since $y e^{y}$ is constant with respect to $x$ , the derivative of $y e^{y}$ with respect to $x$ is $0$ .

$\frac{\partial N}{\partial x} = - (0 + \frac{d}{d x} [- 2 x])$

Step 3.5

Add $0$ and $\frac{d}{d x} [- 2 x]$ .

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

Step 3.6

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

$\frac{\partial N}{\partial x} = - (- 2 \frac{d}{d x} [x])$

Step 3.7

Multiply $- 2$ by $- 1$ .

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

Step 3.8

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

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

Step 3.9

Multiply $2$ by $1$ .

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

Step 4

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

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Step 4.1

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

$1 = 2$

Step 4.2

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

$1 = 2$ is not an identity.

Step 5

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

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Step 5.1

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

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

Step 5.2

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

$\frac{2 - 1}{M}$

Step 5.3

Substitute $y$ for $M$ .

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Step 5.3.1

Substitute $y$ for $M$ .

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

Step 5.3.2

Substitute $y$ for $M$ .

$\frac{1}{y}$

Step 5.4

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

$μ (x, y) = e^{\int \frac{1}{y} d y}$

Step 6

Evaluate the integral $e^{\int \frac{1}{y} d y}$ .

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Step 6.1

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

$μ (x, y) = e^{\ln (| y |) + C}$

Step 6.2

Simplify the answer.

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Step 6.2.1

Simplify.

$μ (x, y) = e^{\ln (| y |)} + C$

Step 6.2.2

Exponentiation and log are inverse functions.

$μ (x, y) = | y | + C$

Step 7

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

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Step 7.1

Multiply $y$ by $y$ .

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

Step 7.2

Multiply $y$ by $y$ .

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

Step 7.3

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

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

Step 7.4

Apply the distributive property.

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

Step 7.5

Multiply $- 2$ by $- 1$ .

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

Step 7.6

Apply the distributive property.

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

Step 7.7

Multiply $y$ by $y$ by adding the exponents.

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Step 7.7.1

Move $y$ .

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

Step 7.7.2

Multiply $y$ by $y$ .

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

Step 8

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

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

Step 9

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

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Step 9.1

Apply the constant rule.

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

Step 10

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

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

Step 11

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

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

Step 12

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

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Step 12.1

Differentiate $f$ with respect to $y$ .

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

Step 12.2

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

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

Step 12.3

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

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Step 12.3.1

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

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

Step 12.3.2

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

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

Step 12.3.3

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

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

Step 12.4

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

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

Step 12.5

Reorder terms.

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

Step 13

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

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Step 13.1

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

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Step 13.1.1

Subtract $2 x y$ from both sides of the equation.

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

Step 13.1.2

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

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Step 13.1.2.1

Subtract $2 x y$ from $2 x y$ .

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

Step 13.1.2.2

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

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

Step 14

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

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Step 14.1

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

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

Step 14.2

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

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

Step 14.3

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

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

Step 14.4

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = y^{2}$ and $d v = e^{y}$ .

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

Step 14.5

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

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

Step 14.6

Multiply $2$ by $- 1$ .

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

Step 14.7

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = y$ and $d v = e^{y}$ .

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

Step 14.8

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

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

Step 14.9

Simplify the answer.

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Step 14.9.1

Rewrite $- (y^{2} e^{y} - 2 (y e^{y} - (e^{y} + C)))$ as $- (y^{2} e^{y} - 2 y e^{y} + 2 e^{y}) + C$ .

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

Step 14.9.2

Simplify.

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Step 14.9.2.1

Apply the distributive property.

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

Step 14.9.2.2

Simplify.

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Step 14.9.2.2.1

Multiply $- 2$ by $- 1$ .

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

Step 14.9.2.2.2

Multiply $2$ by $- 1$ .

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

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

Step 15

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

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