Calculus Examples

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

Step 1

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

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Step 1.1

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

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

Step 2

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

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Step 2.1

Differentiate $M$ with respect to $y$ .

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

Step 2.2

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

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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

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

Step 2.3.3

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

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

Step 2.4

Differentiate.

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Step 2.4.1

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

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

Step 2.4.2

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

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

Step 2.4.3

Add $0$ and $\frac{d}{d y} [y]$ .

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

Step 2.4.4

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

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

Step 2.4.5

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

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

Step 3

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

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Step 3.1

Differentiate $N$ with respect to $x$ .

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

Step 3.2

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

$\frac{\partial N}{\partial x} = 0$

Step 4

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

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Step 4.1

Substitute $x e^{x^{2} + y}$ for $\frac{\partial M}{\partial y}$ and $0$ for $\frac{\partial N}{\partial x}$ .

$x e^{x^{2} + y} = 0$

Step 4.2

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

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

$\frac{0 - \frac{\partial M}{\partial y}}{M}$

Step 5.2

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

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

Step 5.3

Substitute $x e^{x^{2} + y}$ for $M$ .

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Step 5.3.1

Substitute $x e^{x^{2} + y}$ for $M$ .

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

Step 5.3.2

Subtract $x e^{x^{2} + y}$ from $0$ .

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

Step 5.3.3

Cancel the common factor of $x$ .

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Step 5.3.3.1

Cancel the common factor.

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

Step 5.3.3.2

Rewrite the expression.

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

Step 5.3.4

Substitute $x e^{x^{2} + y}$ for $M$ .

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Step 5.3.4.1

Cancel the common factor.

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

Step 5.3.4.2

Divide $- 1$ by $1$ .

$- 1$

Step 5.4

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

$μ (x, y) = e^{\int - 1 d y}$

Step 6

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

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Step 6.1

Apply the constant rule.

$μ (x, y) = e^{- y + C}$

Step 6.2

Simplify.

$μ (x, y) = e^{- y} + C$

Step 7

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

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Step 7.1

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

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

Step 7.2

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

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Step 7.2.1

Move $e^{- y}$ .

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

Step 7.2.2

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

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

Step 7.2.3

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

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Step 7.2.3.1

Add $- y$ and $y$ .

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

Step 7.2.3.2

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

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

Step 7.3

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

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

Step 8

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

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

Step 9

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

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Step 9.1

Let $u_{2} = e^{x^{2}}$ . Then $d u_{2} = 2 x e^{x^{2}} d x$ , so $\frac{1}{2} d u_{2} = x e^{x^{2}} d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 9.1.1

Let $u_{2} = e^{x^{2}}$ . Find $\frac{d u_{2}}{d x}$ .

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Step 9.1.1.1

Differentiate $e^{x^{2}}$ .

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

Step 9.1.1.2

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

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Step 9.1.1.2.1

To apply the Chain Rule, set $u_{1}$ as $x^{2}$ .

$\frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d x} [x^{2}]$

Step 9.1.1.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$e^{u_{1}} \frac{d}{d x} [x^{2}]$

Step 9.1.1.2.3

Replace all occurrences of $u_{1}$ with $x^{2}$ .

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

Step 9.1.1.3

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

$e^{x^{2}} (2 x)$

Step 9.1.1.4

Simplify.

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Step 9.1.1.4.1

Reorder the factors of $e^{x^{2}} (2 x)$ .

$2 e^{x^{2}} x$

Step 9.1.1.4.2

Reorder factors in $2 e^{x^{2}} x$ .

$2 x e^{x^{2}}$

Step 9.1.2

Rewrite the problem using $u_{2}$ and $d u_{2}$ .

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

Step 9.2

Apply the constant rule.

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

Step 9.3

Replace all occurrences of $u_{2}$ with $e^{x^{2}}$ .

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

Step 10

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

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

Step 11

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

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

Step 12

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

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Step 12.1

Differentiate $f$ with respect to $y$ .

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

Step 12.2

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

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

Step 12.3

Since $\frac{1}{2} e^{x^{2}}$ is constant with respect to $y$ , the derivative of $\frac{1}{2} e^{x^{2}}$ with respect to $y$ is $0$ .

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

Step 12.4

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

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

Step 12.5

Add $0$ and $\frac{d g}{d y}$ .

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

Step 13

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

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Step 13.1

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

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

Step 13.2

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

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

Step 13.3

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

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

Step 13.4

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

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

Step 13.5

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

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

Step 13.6

Simplify.

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Step 13.6.1

Multiply $- 1$ by $- 1$ .

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

Step 13.6.2

Multiply $\int e^{- y} d y$ by $1$ .

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

Step 13.7

Let $u = - y$ . Then $d u = - d y$ , so $- d u = d y$ . Rewrite using $u$ and $d$ $u$ .

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Step 13.7.1

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

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Step 13.7.1.1

Differentiate $- y$ .

$\frac{d}{d y} [- y]$

Step 13.7.1.2

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

$- \frac{d}{d y} [y]$

Step 13.7.1.3

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

$- 1 \cdot 1$

Step 13.7.1.4

Multiply $- 1$ by $1$ .

$- 1$

Step 13.7.2

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

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

Step 13.8

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

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

Step 13.9

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

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

Step 13.10

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

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

Step 13.11

Replace all occurrences of $u$ with $- y$ .

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

Step 13.12

Simplify.

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Step 13.12.1

Apply the distributive property.

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

Step 13.12.2

Multiply $- (- y e^{- y})$ .

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Step 13.12.2.1

Multiply $- 1$ by $- 1$ .

$g (y) = 1 (y e^{- y}) - - e^{- y} + C$

Step 13.12.2.2

Multiply $y$ by $1$ .

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

Step 13.12.3

Multiply $- - e^{- y}$ .

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Step 13.12.3.1

Multiply $- 1$ by $- 1$ .

$g (y) = y e^{- y} + 1 e^{- y} + C$

Step 13.12.3.2

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

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

Step 14

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

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

Step 15

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

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