Calculus Examples

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

Step 1

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

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Step 1.1

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

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

Step 1.2

Rewrite.

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

Step 2

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

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Step 2.1

Differentiate $M$ with respect to $y$ .

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

Step 2.2

Differentiate.

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.2.3

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

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

Step 2.2.4

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

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

Step 2.2.5

Multiply $2$ by $1$ .

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

Step 2.2.6

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

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

Step 2.3

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

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

Step 2.4

Differentiate using the Constant Rule.

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Step 2.4.1

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

$\frac{\partial M}{\partial y} = - (2 x - e^{y} + 0)$

Step 2.4.2

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

$\frac{\partial M}{\partial y} = - (2 x - e^{y})$

Step 2.5

Simplify.

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Step 2.5.1

Apply the distributive property.

$\frac{\partial M}{\partial y} = - (2 x) - - e^{y}$

Step 2.5.2

Combine terms.

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Step 2.5.2.1

Multiply $2$ by $- 1$ .

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

Step 2.5.2.2

Multiply $- 1$ by $- 1$ .

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

Step 2.5.2.3

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

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

Step 3

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

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Step 3.1

Differentiate $N$ with respect to $x$ .

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

Step 3.2

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

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

Step 3.3

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

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Step 3.3.1

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

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

Step 3.3.2

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

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

Step 3.3.3

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

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

Step 3.4

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

$\frac{\partial N}{\partial x} = e^{y} + 0 + \frac{d}{d x} [- x^{2}]$

Step 3.5

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

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Step 3.5.1

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

$\frac{\partial N}{\partial x} = e^{y} + 0 - \frac{d}{d x} [x^{2}]$

Step 3.5.2

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

$\frac{\partial N}{\partial x} = e^{y} + 0 - (2 x)$

Step 3.5.3

Multiply $2$ by $- 1$ .

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

Step 3.6

Simplify.

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Step 3.6.1

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

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

Step 3.6.2

Reorder terms.

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

Step 4

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

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Step 4.1

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

$- 2 x + e^{y} = - 2 x + e^{y}$

Step 4.2

Since the two sides have been shown to be equivalent, the equation is an identity.

$- 2 x + e^{y} = - 2 x + e^{y}$ is an identity.

Step 5

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

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

Step 6

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

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Step 6.1

Split the single integral into multiple integrals.

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

Step 6.2

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

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

Step 6.3

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

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

Step 6.4

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

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

Step 6.5

Apply the constant rule.

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

Step 6.6

Simplify.

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

Step 7

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

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

Step 8

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

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

Step 9

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

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Step 9.1

Differentiate $f$ with respect to $x$ .

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

Step 9.2

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

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

Step 9.3

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

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Step 9.3.1

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

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

Step 9.3.2

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

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

Step 9.3.3

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

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

Step 9.4

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 $0$ .

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

Step 9.5

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

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Step 9.5.1

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

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

Step 9.5.2

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

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

Step 9.5.3

Multiply $2$ by $- 1$ .

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

Step 9.6

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

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

Step 9.7

Simplify.

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Step 9.7.1

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

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

Step 9.7.2

Reorder terms.

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

Step 10

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

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Step 10.1

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

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Step 10.1.1

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

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Step 10.1.1.1

Rewrite.

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

Step 10.1.1.2

Simplify by adding zeros.

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

Step 10.1.1.3

Apply the distributive property.

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

Step 10.1.1.4

Simplify.

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Step 10.1.1.4.1

Multiply $2$ by $- 1$ .

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

Step 10.1.1.4.2

Multiply $- - e^{y}$ .

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Step 10.1.1.4.2.1

Multiply $- 1$ by $- 1$ .

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

Step 10.1.1.4.2.2

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

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

Step 10.1.1.4.3

Multiply $- - x$ .

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Step 10.1.1.4.3.1

Multiply $- 1$ by $- 1$ .

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

Step 10.1.1.4.3.2

Multiply $x$ by $1$ .

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

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

Step 10.1.2

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

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Step 10.1.2.1

Add $2 x y$ to both sides of the equation.

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

Step 10.1.2.2

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

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

Step 10.1.2.3

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

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Step 10.1.2.3.1

Add $- 2 x y$ and $2 x y$ .

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

Step 10.1.2.3.2

Add $e^{y} + x$ and $0$ .

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

Step 10.1.2.3.3

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

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

Step 10.1.2.3.4

Add $x$ and $0$ .

$\frac{d g}{d x} = x$

Step 11

Find the antiderivative of $x$ to find $g (x)$ .

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Step 11.1

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

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

Step 11.2

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

$g (x) = \int x d x$

Step 11.3

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

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

Step 12

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

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

Step 13

Simplify each term.

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Step 13.1

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

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

Step 13.2

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

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