Calculus Examples

$\frac{3 d y}{d x} - 18 x = - 6 x y$

Step 1

Separate the variables.

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Step 1.1

Factor $\frac{d y}{d x}$ out.

$3 \frac{d y}{d x} - 18 x = - 6 x y$

Step 1.2

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

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Step 1.2.1

Add $18 x$ to both sides of the equation.

$3 \frac{d y}{d x} = - 6 x y + 18 x$

Step 1.2.2

Divide each term in $3 \frac{d y}{d x} = - 6 x y + 18 x$ by $3$ and simplify.

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Step 1.2.2.1

Divide each term in $3 \frac{d y}{d x} = - 6 x y + 18 x$ by $3$ .

$\frac{3 \frac{d y}{d x}}{3} = \frac{- 6 x y}{3} + \frac{18 x}{3}$

Step 1.2.2.2

Simplify the left side.

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Step 1.2.2.2.1

Cancel the common factor of $3$ .

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Step 1.2.2.2.1.1

Cancel the common factor.

$\frac{3 \frac{d y}{d x}}{3} = \frac{- 6 x y}{3} + \frac{18 x}{3}$

Step 1.2.2.2.1.2

Divide $\frac{d y}{d x}$ by $1$ .

$\frac{d y}{d x} = \frac{- 6 x y}{3} + \frac{18 x}{3}$

Step 1.2.2.3

Simplify the right side.

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Step 1.2.2.3.1

Simplify each term.

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Step 1.2.2.3.1.1

Cancel the common factor of $- 6$ and $3$ .

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Step 1.2.2.3.1.1.1

Factor $3$ out of $- 6 x y$ .

$\frac{d y}{d x} = \frac{3 (- 2 x y)}{3} + \frac{18 x}{3}$

Step 1.2.2.3.1.1.2

Cancel the common factors.

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Step 1.2.2.3.1.1.2.1

Factor $3$ out of $3$ .

$\frac{d y}{d x} = \frac{3 (- 2 x y)}{3 (1)} + \frac{18 x}{3}$

Step 1.2.2.3.1.1.2.2

Cancel the common factor.

$\frac{d y}{d x} = \frac{3 (- 2 x y)}{3 \cdot 1} + \frac{18 x}{3}$

Step 1.2.2.3.1.1.2.3

Rewrite the expression.

$\frac{d y}{d x} = \frac{- 2 x y}{1} + \frac{18 x}{3}$

Step 1.2.2.3.1.1.2.4

Divide $- 2 x y$ by $1$ .

$\frac{d y}{d x} = - 2 x y + \frac{18 x}{3}$

Step 1.2.2.3.1.2

Cancel the common factor of $18$ and $3$ .

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Step 1.2.2.3.1.2.1

Factor $3$ out of $18 x$ .

$\frac{d y}{d x} = - 2 x y + \frac{3 (6 x)}{3}$

Step 1.2.2.3.1.2.2

Cancel the common factors.

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Step 1.2.2.3.1.2.2.1

Factor $3$ out of $3$ .

$\frac{d y}{d x} = - 2 x y + \frac{3 (6 x)}{3 (1)}$

Step 1.2.2.3.1.2.2.2

Cancel the common factor.

$\frac{d y}{d x} = - 2 x y + \frac{3 (6 x)}{3 \cdot 1}$

Step 1.2.2.3.1.2.2.3

Rewrite the expression.

$\frac{d y}{d x} = - 2 x y + \frac{6 x}{1}$

Step 1.2.2.3.1.2.2.4

Divide $6 x$ by $1$ .

$\frac{d y}{d x} = - 2 x y + 6 x$

Step 1.3

Factor.

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Step 1.3.1

Factor $2 x$ out of $- 2 x y + 6 x$ .

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Step 1.3.1.1

Factor $2 x$ out of $- 2 x y$ .

$\frac{d y}{d x} = 2 x (- 1 y) + 6 x$

Step 1.3.1.2

Factor $2 x$ out of $6 x$ .

$\frac{d y}{d x} = 2 x (- 1 y) + 2 x (3)$

Step 1.3.1.3

Factor $2 x$ out of $2 x (- 1 y) + 2 x (3)$ .

$\frac{d y}{d x} = 2 x (- 1 y + 3)$

Step 1.3.2

Rewrite $- 1 y$ as $- y$ .

$\frac{d y}{d x} = 2 x (- y + 3)$

Step 1.4

Multiply both sides by $\frac{1}{- y + 3}$ .

$\frac{1}{- y + 3} \frac{d y}{d x} = \frac{1}{- y + 3} (2 x (- y + 3))$

Step 1.5

Simplify.

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Step 1.5.1

Rewrite using the commutative property of multiplication.

$\frac{1}{- y + 3} \frac{d y}{d x} = 2 \frac{1}{- y + 3} (x (- y + 3))$

Step 1.5.2

Combine $2$ and $\frac{1}{- y + 3}$ .

$\frac{1}{- y + 3} \frac{d y}{d x} = \frac{2}{- y + 3} (x (- y + 3))$

Step 1.5.3

Cancel the common factor of $- y + 3$ .

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Step 1.5.3.1

Factor $- y + 3$ out of $x (- y + 3)$ .

$\frac{1}{- y + 3} \frac{d y}{d x} = \frac{2}{- y + 3} ((- y + 3) x)$

Step 1.5.3.2

Cancel the common factor.

$\frac{1}{- y + 3} \frac{d y}{d x} = \frac{2}{- y + 3} ((- y + 3) x)$

Step 1.5.3.3

Rewrite the expression.

$\frac{1}{- y + 3} \frac{d y}{d x} = 2 x$

Step 1.6

Rewrite the equation.

$\frac{1}{- y + 3} d y = 2 x d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1}{- y + 3} d y = \int 2 x d x$

Step 2.2

Integrate the left side.

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Step 2.2.1

Let $u = - y + 3$ . Then $d u = - d y$ , so $- d u = d y$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.2.1.1

Let $u = - y + 3$ . Find $\frac{d u}{d y}$ .

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Step 2.2.1.1.1

Rewrite.

$\frac{- 1}{1}$

Step 2.2.1.1.2

Divide $- 1$ by $1$ .

$- 1$

Step 2.2.1.2

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

$\int \frac{- 1}{u} d u = \int 2 x d x$

Step 2.2.2

Split the fraction into multiple fractions.

$\int - \frac{1}{u} d u = \int 2 x d x$

Step 2.2.3

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

$- \int \frac{1}{u} d u = \int 2 x d x$

Step 2.2.4

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

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

Step 2.2.5

Simplify.

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

Step 2.2.6

Replace all occurrences of $u$ with $- y + 3$ .

$- \ln (| - y + 3 |) + C_{1} = \int 2 x d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Since $2$ is constant with respect to $x$ , move $2$ out of the integral.

$- \ln (| - y + 3 |) + C_{1} = 2 \int x d x$

Step 2.3.2

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

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

Step 2.3.3

Simplify the answer.

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Step 2.3.3.1

Rewrite $2 (\frac{1}{2} x^{2} + C_{2})$ as $2 (\frac{1}{2}) x^{2} + C_{2}$ .

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

Step 2.3.3.2

Simplify.

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Step 2.3.3.2.1

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

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

Step 2.3.3.2.2

Cancel the common factor of $2$ .

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Step 2.3.3.2.2.1

Cancel the common factor.

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

Step 2.3.3.2.2.2

Rewrite the expression.

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

Step 2.3.3.2.3

Multiply $x^{2}$ by $1$ .

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

Step 2.4

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

$- \ln (| - y + 3 |) = x^{2} + C$

Step 3

Solve for $y$ .

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Step 3.1

Divide each term in $- \ln (| - y + 3 |) = x^{2} + C$ by $- 1$ and simplify.

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Step 3.1.1

Divide each term in $- \ln (| - y + 3 |) = x^{2} + C$ by $- 1$ .

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

Step 3.1.2

Simplify the left side.

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Step 3.1.2.1

Dividing two negative values results in a positive value.

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

Step 3.1.2.2

Divide $\ln (| - y + 3 |)$ by $1$ .

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

Step 3.1.3

Simplify the right side.

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Step 3.1.3.1

Simplify each term.

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Step 3.1.3.1.1

Move the negative one from the denominator of $\frac{x^{2}}{- 1}$ .

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

Step 3.1.3.1.2

Rewrite $- 1 \cdot x^{2}$ as $- x^{2}$ .

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

Step 3.1.3.1.3

Move the negative one from the denominator of $\frac{C}{- 1}$ .

$\ln (| - y + 3 |) = - x^{2} - 1 \cdot C$

Step 3.1.3.1.4

Rewrite $- 1 \cdot C$ as $- C$ .

$\ln (| - y + 3 |) = - x^{2} - C$

Step 3.2

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

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

Step 3.3

Rewrite $\ln (| - y + 3 |) = - x^{2} - C$ 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 $b^{y} = x$ .

$e^{- x^{2} - C} = | - y + 3 |$

Step 3.4

Solve for $y$ .

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Step 3.4.1

Rewrite the equation as $| - y + 3 | = e^{- x^{2} - C}$ .

$| - y + 3 | = e^{- x^{2} - C}$

Step 3.4.2

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

$- y + 3 = \pm e^{- x^{2} - C}$

Step 3.4.3

Subtract $3$ from both sides of the equation.

$- y = \pm e^{- x^{2} - C} - 3$

Step 3.4.4

Divide each term in $- y = \pm e^{- x^{2} - C} - 3$ by $- 1$ and simplify.

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Step 3.4.4.1

Divide each term in $- y = \pm e^{- x^{2} - C} - 3$ by $- 1$ .

$\frac{- y}{- 1} = \frac{\pm e^{- x^{2} - C}}{- 1} + \frac{- 3}{- 1}$

Step 3.4.4.2

Simplify the left side.

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Step 3.4.4.2.1

Dividing two negative values results in a positive value.

$\frac{y}{1} = \frac{\pm e^{- x^{2} - C}}{- 1} + \frac{- 3}{- 1}$

Step 3.4.4.2.2

Divide $y$ by $1$ .

$y = \frac{\pm e^{- x^{2} - C}}{- 1} + \frac{- 3}{- 1}$

Step 3.4.4.3

Simplify the right side.

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Step 3.4.4.3.1

Simplify each term.

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Step 3.4.4.3.1.1

Move the negative one from the denominator of $\frac{\pm e^{- x^{2} - C}}{- 1}$ .

$y = - 1 \cdot (\pm e^{- x^{2} - C}) + \frac{- 3}{- 1}$

Step 3.4.4.3.1.2

Rewrite $- 1 \cdot (\pm e^{- x^{2} - C})$ as $- (\pm e^{- x^{2} - C})$ .

$y = - (\pm e^{- x^{2} - C}) + \frac{- 3}{- 1}$

Step 3.4.4.3.1.3

Divide $- 3$ by $- 1$ .

$y = - (\pm e^{- x^{2} - C}) + 3$

Step 4

Group the constant terms together.

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Step 4.1

Simplify the constant of integration.

$y = - (\pm e^{- x^{2} + C}) + 3$

Step 4.2

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

$y = - (\pm e^{- x^{2}} e^{C}) + 3$

Step 4.3

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

$y = - (\pm e^{C} e^{- x^{2}}) + 3$

Step 4.4

Combine constants with the plus or minus.

$y = - (C e^{- x^{2}}) + 3$