Calculus Examples

$\frac{d y}{d x} = 6 x^{2} y - 5 x y$

Step 1

Separate the variables.

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Step 1.1

Factor $x y$ out of $6 x^{2} y - 5 x y$ .

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Step 1.1.1

Factor $x y$ out of $6 x^{2} y$ .

$\frac{d y}{d x} = x y (6 x) - 5 x y$

Step 1.1.2

Factor $x y$ out of $- 5 x y$ .

$\frac{d y}{d x} = x y (6 x) + x y (- 5)$

Step 1.1.3

Factor $x y$ out of $x y (6 x) + x y (- 5)$ .

$\frac{d y}{d x} = x y (6 x - 5)$

Step 1.2

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

$\frac{1}{y} \frac{d y}{d x} = \frac{1}{y} (x y (6 x - 5))$

Step 1.3

Simplify.

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Step 1.3.1

Cancel the common factor of $y$ .

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Step 1.3.1.1

Factor $y$ out of $x y (6 x - 5)$ .

$\frac{1}{y} \frac{d y}{d x} = \frac{1}{y} (y (x (6 x - 5)))$

Step 1.3.1.2

Cancel the common factor.

$\frac{1}{y} \frac{d y}{d x} = \frac{1}{y} (y (x (6 x - 5)))$

Step 1.3.1.3

Rewrite the expression.

$\frac{1}{y} \frac{d y}{d x} = x (6 x - 5)$

Step 1.3.2

Apply the distributive property.

$\frac{1}{y} \frac{d y}{d x} = x (6 x) + x \cdot - 5$

Step 1.3.3

Rewrite using the commutative property of multiplication.

$\frac{1}{y} \frac{d y}{d x} = 6 x \cdot x + x \cdot - 5$

Step 1.3.4

Move $- 5$ to the left of $x$ .

$\frac{1}{y} \frac{d y}{d x} = 6 x \cdot x - 5 \cdot x$

Step 1.3.5

Multiply $x$ by $x$ by adding the exponents.

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Step 1.3.5.1

Move $x$ .

$\frac{1}{y} \frac{d y}{d x} = 6 (x \cdot x) - 5 \cdot x$

Step 1.3.5.2

Multiply $x$ by $x$ .

$\frac{1}{y} \frac{d y}{d x} = 6 x^{2} - 5 \cdot x$

$\frac{1}{y} \frac{d y}{d x} = 6 x^{2} - 5 x$

Step 1.4

Rewrite the equation.

$\frac{1}{y} d y = (6 x^{2} - 5 x) d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1}{y} d y = \int 6 x^{2} - 5 x d x$

Step 2.2

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

$\ln (| y |) + C_{1} = \int 6 x^{2} - 5 x d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

$\ln (| y |) + C_{1} = \int 6 x^{2} d x + \int - 5 x d x$

Step 2.3.2

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

$\ln (| y |) + C_{1} = 6 \int x^{2} d x + \int - 5 x d x$

Step 2.3.3

By the Power Rule, the integral of $x^{2}$ with respect to $x$ is $\frac{1}{3} x^{3}$ .

$\ln (| y |) + C_{1} = 6 (\frac{1}{3} x^{3} + C_{2}) + \int - 5 x d x$

Step 2.3.4

Since $- 5$ is constant with respect to $x$ , move $- 5$ out of the integral.

$\ln (| y |) + C_{1} = 6 (\frac{1}{3} x^{3} + C_{2}) - 5 \int x d x$

Step 2.3.5

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

$\ln (| y |) + C_{1} = 6 (\frac{1}{3} x^{3} + C_{2}) - 5 (\frac{1}{2} x^{2} + C_{3})$

Step 2.3.6

Simplify.

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Step 2.3.6.1

Simplify.

$\ln (| y |) + C_{1} = 6 (\frac{1}{3}) x^{3} - 5 (\frac{1}{2}) x^{2} + C_{4}$

Step 2.3.6.2

Simplify.

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Step 2.3.6.2.1

Combine $6$ and $\frac{1}{3}$ .

$\ln (| y |) + C_{1} = \frac{6}{3} x^{3} - 5 (\frac{1}{2}) x^{2} + C_{4}$

Step 2.3.6.2.2

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

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Step 2.3.6.2.2.1

Factor $3$ out of $6$ .

$\ln (| y |) + C_{1} = \frac{3 \cdot 2}{3} x^{3} - 5 (\frac{1}{2}) x^{2} + C_{4}$

Step 2.3.6.2.2.2

Cancel the common factors.

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Step 2.3.6.2.2.2.1

Factor $3$ out of $3$ .

$\ln (| y |) + C_{1} = \frac{3 \cdot 2}{3 (1)} x^{3} - 5 (\frac{1}{2}) x^{2} + C_{4}$

Step 2.3.6.2.2.2.2

Cancel the common factor.

$\ln (| y |) + C_{1} = \frac{3 \cdot 2}{3 \cdot 1} x^{3} - 5 (\frac{1}{2}) x^{2} + C_{4}$

Step 2.3.6.2.2.2.3

Rewrite the expression.

$\ln (| y |) + C_{1} = \frac{2}{1} x^{3} - 5 (\frac{1}{2}) x^{2} + C_{4}$

Step 2.3.6.2.2.2.4

Divide $2$ by $1$ .

$\ln (| y |) + C_{1} = 2 x^{3} - 5 (\frac{1}{2}) x^{2} + C_{4}$

Step 2.3.6.2.3

Combine $- 5$ and $\frac{1}{2}$ .

$\ln (| y |) + C_{1} = 2 x^{3} + \frac{- 5}{2} x^{2} + C_{4}$

Step 2.3.6.2.4

Move the negative in front of the fraction.

$\ln (| y |) + C_{1} = 2 x^{3} - \frac{5}{2} x^{2} + C_{4}$

Step 2.4

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

$\ln (| y |) = 2 x^{3} - \frac{5}{2} x^{2} + C$

Step 3

Solve for $y$ .

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Step 3.1

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

$e^{\ln (| y |)} = e^{2 x^{3} - \frac{5}{2} x^{2} + C}$

Step 3.2

Rewrite $\ln (| y |) = 2 x^{3} - \frac{5}{2} x^{2} + C$ 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 $b^{y} = x$ .

$e^{2 x^{3} - \frac{5}{2} x^{2} + C} = | y |$

Step 3.3

Solve for $y$ .

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Step 3.3.1

Rewrite the equation as $| y | = e^{2 x^{3} - \frac{5}{2} x^{2} + C}$ .

$| y | = e^{2 x^{3} - \frac{5}{2} x^{2} + C}$

Step 3.3.2

Simplify each term.

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Step 3.3.2.1

Combine $x^{2}$ and $\frac{5}{2}$ .

$| y | = e^{2 x^{3} - \frac{x^{2} \cdot 5}{2} + C}$

Step 3.3.2.2

Move $5$ to the left of $x^{2}$ .

$| y | = e^{2 x^{3} - \frac{5 x^{2}}{2} + C}$

Step 3.3.3

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

$y = \pm e^{2 x^{3} - \frac{5 x^{2}}{2} + C}$

Step 4

Group the constant terms together.

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Step 4.1

Rewrite $e^{2 x^{3} - \frac{5 x^{2}}{2} + C}$ as $e^{2 x^{3} - \frac{5 x^{2}}{2}} e^{C}$ .

$y = \pm e^{2 x^{3} - \frac{5 x^{2}}{2}} e^{C}$

Step 4.2

Reorder $e^{2 x^{3} - \frac{5 x^{2}}{2}}$ and $e^{C}$ .

$y = \pm e^{C} e^{2 x^{3} - \frac{5 x^{2}}{2}}$

Step 4.3

Combine constants with the plus or minus.

$y = C e^{2 x^{3} - \frac{5 x^{2}}{2}}$