Calculus Examples

$y (x + 9) + \frac{d y}{d x} = 0$

Step 1

Separate the variables.

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Step 1.1

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

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Step 1.1.1

Simplify each term.

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Step 1.1.1.1

Apply the distributive property.

$y x + y \cdot 9 + \frac{d y}{d x} = 0$

Step 1.1.1.2

Move $9$ to the left of $y$ .

$y x + 9 y + \frac{d y}{d x} = 0$

Step 1.1.2

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

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Step 1.1.2.1

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

$9 y + \frac{d y}{d x} = - y x$

Step 1.1.2.2

Subtract $9 y$ from both sides of the equation.

$\frac{d y}{d x} = - y x - 9 y$

Step 1.2

Factor.

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Step 1.2.1

Factor $y$ out of $- y x - 9 y$ .

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Step 1.2.1.1

Factor $y$ out of $- y x$ .

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

Step 1.2.1.2

Factor $y$ out of $- 9 y$ .

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

Step 1.2.1.3

Factor $y$ out of $y (- 1 x) + y \cdot - 9$ .

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

Step 1.2.2

Rewrite $- 1 x$ as $- x$ .

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

Step 1.3

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

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

Step 1.4

Cancel the common factor of $y$ .

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Step 1.4.1

Cancel the common factor.

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

Step 1.4.2

Rewrite the expression.

$\frac{1}{y} \frac{d y}{d x} = - x - 9$

Step 1.5

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

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

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

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

Step 2.3.2

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

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

Step 2.3.3

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

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

Step 2.3.4

Apply the constant rule.

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

Step 2.3.5

Simplify.

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Step 2.3.5.1

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

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

Step 2.3.5.2

Simplify.

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

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

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

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

Step 3.2

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

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

Step 3.3

Solve for $y$ .

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Step 3.3.1

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

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

Step 3.3.2

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

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

Step 3.3.3

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

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

Step 4

Group the constant terms together.

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Step 4.1

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

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

Step 4.2

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

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

Step 4.3

Combine constants with the plus or minus.

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