Calculus Examples

$\frac{d y}{d x} - 6 x y - 12 x = 0$

Step 1

Separate the variables.

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Step 1.1

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

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Step 1.1.1

Add $6 x y$ to both sides of the equation.

$\frac{d y}{d x} - 12 x = 6 x y$

Step 1.1.2

Add $12 x$ to both sides of the equation.

$\frac{d y}{d x} = 6 x y + 12 x$

Step 1.2

Factor $6 x$ out of $6 x y + 12 x$ .

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Step 1.2.1

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

$\frac{d y}{d x} = 6 x (y) + 12 x$

Step 1.2.2

Factor $6 x$ out of $12 x$ .

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

Step 1.2.3

Factor $6 x$ out of $6 x (y) + 6 x (2)$ .

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

Step 1.3

Multiply both sides by $\frac{1}{y + 2}$ .

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

Step 1.4

Simplify.

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Step 1.4.1

Rewrite using the commutative property of multiplication.

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

Step 1.4.2

Combine $6$ and $\frac{1}{y + 2}$ .

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

Step 1.4.3

Cancel the common factor of $y + 2$ .

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Step 1.4.3.1

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

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

Step 1.4.3.2

Cancel the common factor.

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

Step 1.4.3.3

Rewrite the expression.

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

Step 1.5

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

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

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Step 2.2.1.1

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

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Step 2.2.1.1.1

Differentiate $y + 2$ .

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

Step 2.2.1.1.2

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

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

Step 2.2.1.1.3

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

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

Step 2.2.1.1.4

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

$1 + 0$

Step 2.2.1.1.5

Add $1$ and $0$ .

$1$

Step 2.2.1.2

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

$\int \frac{1}{u} d u = \int 6 x d x$

Step 2.2.2

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

$\ln (| u |) + C_{1} = \int 6 x d x$

Step 2.2.3

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

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Since $6$ is constant with respect to $x$ , move $6$ out of the integral.

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

Step 2.3.2

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

$\ln (| y + 2 |) + C_{1} = 6 (\frac{1}{2} x^{2} + C_{2})$

Step 2.3.3

Simplify the answer.

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Step 2.3.3.1

Rewrite $6 (\frac{1}{2} x^{2} + C_{2})$ as $6 (\frac{1}{2}) x^{2} + C_{2}$ .

$\ln (| y + 2 |) + C_{1} = 6 (\frac{1}{2}) x^{2} + C_{2}$

Step 2.3.3.2

Simplify.

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Step 2.3.3.2.1

Combine $6$ and $\frac{1}{2}$ .

$\ln (| y + 2 |) + C_{1} = \frac{6}{2} x^{2} + C_{2}$

Step 2.3.3.2.2

Cancel the common factor of $6$ and $2$ .

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Step 2.3.3.2.2.1

Factor $2$ out of $6$ .

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

Step 2.3.3.2.2.2

Cancel the common factors.

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Step 2.3.3.2.2.2.1

Factor $2$ out of $2$ .

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

Step 2.3.3.2.2.2.2

Cancel the common factor.

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

Step 2.3.3.2.2.2.3

Rewrite the expression.

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

Step 2.3.3.2.2.2.4

Divide $3$ by $1$ .

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

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

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

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

Step 3.2

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

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

Step 3.3

Solve for $y$ .

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Step 3.3.1

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

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

Step 3.3.2

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

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

Step 3.3.3

Subtract $2$ from both sides of the equation.

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

Step 4

Group the constant terms together.

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Step 4.1

Rewrite $e^{3 x^{2} + C}$ as $e^{3 x^{2}} e^{C}$ .

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

Step 4.2

Reorder $e^{3 x^{2}}$ and $e^{C}$ .

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

Step 4.3

Combine constants with the plus or minus.

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