Calculus Examples

$\frac{d y}{d x} = (9 + x) (8 + y)$

Step 1

Separate the variables.

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Step 1.1

Multiply both sides by $\frac{1}{8 + y}$ .

$\frac{1}{8 + y} \frac{d y}{d x} = \frac{1}{8 + y} ((9 + x) (8 + y))$

Step 1.2

Cancel the common factor of $8 + y$ .

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Step 1.2.1

Factor $8 + y$ out of $(9 + x) (8 + y)$ .

$\frac{1}{8 + y} \frac{d y}{d x} = \frac{1}{8 + y} ((8 + y) (9 + x))$

Step 1.2.2

Cancel the common factor.

$\frac{1}{8 + y} \frac{d y}{d x} = \frac{1}{8 + y} ((8 + y) (9 + x))$

Step 1.2.3

Rewrite the expression.

$\frac{1}{8 + y} \frac{d y}{d x} = 9 + x$

Step 1.3

Rewrite the equation.

$\frac{1}{8 + y} d y = (9 + x) d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1}{8 + y} d y = \int 9 + x d x$

Step 2.2

Integrate the left side.

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Step 2.2.1

Let $u = 8 + y$ . Then $d u = d y$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.2.1.1

Let $u = 8 + y$ . Find $\frac{d u}{d y}$ .

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Step 2.2.1.1.1

Differentiate $8 + y$ .

$\frac{d}{d y} [8 + y]$

Step 2.2.1.1.2

By the Sum Rule, the derivative of $8 + y$ with respect to $y$ is $\frac{d}{d y} [8] + \frac{d}{d y} [y]$ .

$\frac{d}{d y} [8] + \frac{d}{d y} [y]$

Step 2.2.1.1.3

Since $8$ is constant with respect to $y$ , the derivative of $8$ with respect to $y$ is $0$ .

$0 + \frac{d}{d y} [y]$

Step 2.2.1.1.4

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

$0 + 1$

Step 2.2.1.1.5

Add $0$ and $1$ .

$1$

Step 2.2.1.2

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

$\int \frac{1}{u} d u = \int 9 + x d x$

Step 2.2.2

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

$\ln (| u |) + C_{1} = \int 9 + x d x$

Step 2.2.3

Replace all occurrences of $u$ with $8 + y$ .

$\ln (| 8 + y |) + C_{1} = \int 9 + x d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

$\ln (| 8 + y |) + C_{1} = \int 9 d x + \int x d x$

Step 2.3.2

Apply the constant rule.

$\ln (| 8 + y |) + C_{1} = 9 x + C_{2} + \int x d x$

Step 2.3.3

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

$\ln (| 8 + y |) + C_{1} = 9 x + C_{2} + \frac{1}{2} x^{2} + C_{3}$

Step 2.3.4

Simplify.

$\ln (| 8 + y |) + C_{1} = 9 x + \frac{1}{2} x^{2} + C_{4}$

Step 2.3.5

Reorder terms.

$\ln (| 8 + y |) + C_{1} = \frac{1}{2} x^{2} + 9 x + C_{4}$

Step 2.4

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

$\ln (| 8 + y |) = \frac{1}{2} x^{2} + 9 x + C$

Step 3

Solve for $y$ .

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Step 3.1

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

$e^{\ln (| 8 + y |)} = e^{\frac{1}{2} x^{2} + 9 x + C}$

Step 3.2

Rewrite $\ln (| 8 + y |) = \frac{1}{2} x^{2} + 9 x + C$ 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 $b^{y} = x$ .

$e^{\frac{1}{2} x^{2} + 9 x + C} = | 8 + y |$

Step 3.3

Solve for $y$ .

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Step 3.3.1

Rewrite the equation as $| 8 + y | = e^{\frac{1}{2} x^{2} + 9 x + C}$ .

$| 8 + y | = e^{\frac{1}{2} x^{2} + 9 x + C}$

Step 3.3.2

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

$| 8 + y | = e^{\frac{x^{2}}{2} + 9 x + C}$

Step 3.3.3

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

$8 + y = \pm e^{\frac{x^{2}}{2} + 9 x + C}$

Step 3.3.4

Subtract $8$ from both sides of the equation.

$y = \pm e^{\frac{x^{2}}{2} + 9 x + C} - 8$

Step 4

Group the constant terms together.

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Step 4.1

Rewrite $e^{\frac{x^{2}}{2} + 9 x + C}$ as $e^{\frac{x^{2}}{2} + 9 x} e^{C}$ .

$y = \pm e^{\frac{x^{2}}{2} + 9 x} e^{C} - 8$

Step 4.2

Reorder $e^{\frac{x^{2}}{2} + 9 x}$ and $e^{C}$ .

$y = \pm e^{C} e^{\frac{x^{2}}{2} + 9 x} - 8$

Step 4.3

Combine constants with the plus or minus.

$y = C e^{\frac{x^{2}}{2} + 9 x} - 8$