Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

Subtract $y$ from both sides of the equation.

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

Step 1.2

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

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Step 1.2.1

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

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

Step 1.2.2

Factor $y$ out of $- y$ .

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

Step 1.2.3

Factor $y$ out of $y (x e^{x + 2}) + y \cdot - 1$ .

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

Step 1.3

Multiply both sides by $\frac{1}{y}$ .

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

Step 1.4

Cancel the common factor of $y$ .

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Step 1.4.1

Cancel the common factor.

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

Step 1.4.2

Rewrite the expression.

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

Step 1.5

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1}{y} d y = \int x e^{x + 2} - 1 d x$

Step 2.2

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

$\ln (| y |) + C_{1} = \int x e^{x + 2} - 1 d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

$\ln (| y |) + C_{1} = \int x e^{x + 2} d x + \int - 1 d x$

Step 2.3.2

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

$\ln (| y |) + C_{1} = x e^{x + 2} - \int e^{x + 2} d x + \int - 1 d x$

Step 2.3.3

Let $u = x + 2$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.3.3.1

Let $u = x + 2$ . Find $\frac{d u}{d x}$ .

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Step 2.3.3.1.1

Differentiate $x + 2$ .

$\frac{d}{d x} [x + 2]$

Step 2.3.3.1.2

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

$\frac{d}{d x} [x] + \frac{d}{d x} [2]$

Step 2.3.3.1.3

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

$1 + \frac{d}{d x} [2]$

Step 2.3.3.1.4

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

$1 + 0$

Step 2.3.3.1.5

Add $1$ and $0$ .

$1$

Step 2.3.3.2

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

$\ln (| y |) + C_{1} = x e^{x + 2} - \int e^{u} d u + \int - 1 d x$

Step 2.3.4

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

$\ln (| y |) + C_{1} = x e^{x + 2} - (e^{u} + C_{2}) + \int - 1 d x$

Step 2.3.5

Apply the constant rule.

$\ln (| y |) + C_{1} = x e^{x + 2} - (e^{u} + C_{2}) - x + C_{3}$

Step 2.3.6

Simplify.

$\ln (| y |) + C_{1} = x e^{x + 2} - e^{u} - x + C_{4}$

Step 2.3.7

Replace all occurrences of $u$ with $x + 2$ .

$\ln (| y |) + C_{1} = x e^{x + 2} - e^{x + 2} - x + C_{4}$

Step 2.3.8

Reorder terms.

$\ln (| y |) + C_{1} = e^{x + 2} x - x - e^{x + 2} + C_{4}$

Step 2.4

Group the constant of integration on the right side as $C$ .

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

Step 3

Solve for $y$ .

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Step 3.1

To solve for $y$ , rewrite the equation using properties of logarithms.

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

Step 3.2

Rewrite $\ln (| y |) = e^{x + 2} x - x - e^{x + 2} + C$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

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

Step 3.3

Solve for $y$ .

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Step 3.3.1

Rewrite the equation as $| y | = e^{e^{x + 2} x - x - e^{x + 2} + C}$ .

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

Step 3.3.2

Reorder factors in $e^{e^{x + 2} x - x - e^{x + 2} + C}$ .

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

Step 3.3.3

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$y = \pm e^{x e^{x + 2} - x - e^{x + 2} + C}$

Step 4

Group the constant terms together.

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Step 4.1

Rewrite $e^{x e^{x + 2} - x - e^{x + 2} + C}$ as $e^{x e^{x + 2} - x - e^{x + 2}} e^{C}$ .

$y = \pm e^{x e^{x + 2} - x - e^{x + 2}} e^{C}$

Step 4.2

Reorder $e^{x e^{x + 2} - x - e^{x + 2}}$ and $e^{C}$ .

$y = \pm e^{C} e^{x e^{x + 2} - x - e^{x + 2}}$

Step 4.3

Combine constants with the plus or minus.

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