Calculus Examples

$\frac{d x}{d t} = - 6 x + 4 x y$

Step 1

Separate the variables.

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Step 1.1

Factor $2 x$ out of $- 6 x + 4 x y$ .

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Step 1.1.1

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

$\frac{d x}{d t} = 2 x (- 3) + 4 x y$

Step 1.1.2

Factor $2 x$ out of $4 x y$ .

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

Step 1.1.3

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

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

Step 1.2

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

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

Step 1.3

Simplify.

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Step 1.3.1

Rewrite using the commutative property of multiplication.

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

Step 1.3.2

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

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

Step 1.3.3

Cancel the common factor of $x$ .

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Step 1.3.3.1

Cancel the common factor.

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

Step 1.3.3.2

Rewrite the expression.

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

Step 1.3.4

Apply the distributive property.

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

Step 1.3.5

Multiply $2$ by $- 3$ .

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

Step 1.3.6

Multiply $2$ by $2$ .

$\frac{1}{x} \frac{d x}{d t} = - 6 + 4 y$

Step 1.4

Rewrite the equation.

$\frac{1}{x} d x = (- 6 + 4 y) d t$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1}{x} d x = \int - 6 + 4 y d t$

Step 2.2

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

$\ln (| x |) + C_{1} = \int - 6 + 4 y d t$

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

$\ln (| x |) + C_{1} = \int - 6 d t + \int 4 y d t$

Step 2.3.2

Apply the constant rule.

$\ln (| x |) + C_{1} = - 6 t + C_{2} + \int 4 y d t$

Step 2.3.3

Apply the constant rule.

$\ln (| x |) + C_{1} = - 6 t + C_{2} + 4 y t + C_{3}$

Step 2.3.4

Simplify.

$\ln (| x |) + C_{1} = - 6 t + 4 y t + C_{4}$

Step 2.3.5

Reorder terms.

$\ln (| x |) + C_{1} = 4 t y - 6 t + C_{4}$

Step 2.4

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

$\ln (| x |) = 4 t y - 6 t + C$

Step 3

Solve for $x$ .

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Step 3.1

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

$e^{\ln (| x |)} = e^{4 t y - 6 t + C}$

Step 3.2

Rewrite $\ln (| x |) = 4 t y - 6 t + C$ 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 $b^{y} = x$ .

$e^{4 t y - 6 t + C} = | x |$

Step 3.3

Solve for $x$ .

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Step 3.3.1

Rewrite the equation as $| x | = e^{4 t y - 6 t + C}$ .

$| x | = e^{4 t y - 6 t + C}$

Step 3.3.2

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

$x = \pm e^{4 t y - 6 t + C}$

Step 4

Group the constant terms together.

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Step 4.1

Rewrite $e^{4 t y - 6 t + C}$ as $e^{4 t y - 6 t} e^{C}$ .

$x = \pm e^{4 t y - 6 t} e^{C}$

Step 4.2

Reorder $e^{4 t y - 6 t}$ and $e^{C}$ .

$x = \pm e^{C} e^{4 t y - 6 t}$

Step 4.3

Combine constants with the plus or minus.

$x = C e^{4 t y - 6 t}$