Calculus Examples

$\frac{d y}{d x} = 8 x^{2} y - 9 x y$

Step 1

Separate the variables.

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Step 1.1

Factor $x y$ out of $8 x^{2} y - 9 x y$ .

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Step 1.1.1

Factor $x y$ out of $8 x^{2} y$ .

$\frac{d y}{d x} = x y (8 x) - 9 x y$

Step 1.1.2

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

$\frac{d y}{d x} = x y (8 x) + x y (- 9)$

Step 1.1.3

Factor $x y$ out of $x y (8 x) + x y (- 9)$ .

$\frac{d y}{d x} = x y (8 x - 9)$

Step 1.2

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

$\frac{1}{y} \frac{d y}{d x} = \frac{1}{y} (x y (8 x - 9))$

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 $x y (8 x - 9)$ .

$\frac{1}{y} \frac{d y}{d x} = \frac{1}{y} (y (x (8 x - 9)))$

Step 1.3.1.2

Cancel the common factor.

$\frac{1}{y} \frac{d y}{d x} = \frac{1}{y} (y (x (8 x - 9)))$

Step 1.3.1.3

Rewrite the expression.

$\frac{1}{y} \frac{d y}{d x} = x (8 x - 9)$

Step 1.3.2

Apply the distributive property.

$\frac{1}{y} \frac{d y}{d x} = x (8 x) + x \cdot - 9$

Step 1.3.3

Rewrite using the commutative property of multiplication.

$\frac{1}{y} \frac{d y}{d x} = 8 x \cdot x + x \cdot - 9$

Step 1.3.4

Move $- 9$ to the left of $x$ .

$\frac{1}{y} \frac{d y}{d x} = 8 x \cdot x - 9 \cdot x$

Step 1.3.5

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

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Step 1.3.5.1

Move $x$ .

$\frac{1}{y} \frac{d y}{d x} = 8 (x \cdot x) - 9 \cdot x$

Step 1.3.5.2

Multiply $x$ by $x$ .

$\frac{1}{y} \frac{d y}{d x} = 8 x^{2} - 9 \cdot x$

$\frac{1}{y} \frac{d y}{d x} = 8 x^{2} - 9 x$

Step 1.4

Rewrite the equation.

$\frac{1}{y} d y = (8 x^{2} - 9 x) d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1}{y} d y = \int 8 x^{2} - 9 x d x$

Step 2.2

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

$\ln (| y |) + C_{1} = \int 8 x^{2} - 9 x d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

$\ln (| y |) + C_{1} = \int 8 x^{2} d x + \int - 9 x d x$

Step 2.3.2

Since $8$ is constant with respect to $x$ , move $8$ out of the integral.

$\ln (| y |) + C_{1} = 8 \int x^{2} d x + \int - 9 x d x$

Step 2.3.3

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

$\ln (| y |) + C_{1} = 8 (\frac{1}{3} x^{3} + C_{2}) + \int - 9 x d x$

Step 2.3.4

Since $- 9$ is constant with respect to $x$ , move $- 9$ out of the integral.

$\ln (| y |) + C_{1} = 8 (\frac{1}{3} x^{3} + C_{2}) - 9 \int x d x$

Step 2.3.5

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

$\ln (| y |) + C_{1} = 8 (\frac{1}{3} x^{3} + C_{2}) - 9 (\frac{1}{2} x^{2} + C_{3})$

Step 2.3.6

Simplify.

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Step 2.3.6.1

Simplify.

$\ln (| y |) + C_{1} = 8 (\frac{1}{3}) x^{3} - 9 (\frac{1}{2}) x^{2} + C_{4}$

Step 2.3.6.2

Simplify.

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Step 2.3.6.2.1

Combine $8$ and $\frac{1}{3}$ .

$\ln (| y |) + C_{1} = \frac{8}{3} x^{3} - 9 (\frac{1}{2}) x^{2} + C_{4}$

Step 2.3.6.2.2

Combine $- 9$ and $\frac{1}{2}$ .

$\ln (| y |) + C_{1} = \frac{8}{3} x^{3} + \frac{- 9}{2} x^{2} + C_{4}$

Step 2.3.6.2.3

Move the negative in front of the fraction.

$\ln (| y |) + C_{1} = \frac{8}{3} x^{3} - \frac{9}{2} x^{2} + C_{4}$

Step 2.4

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

$\ln (| y |) = \frac{8}{3} x^{3} - \frac{9}{2} x^{2} + C$

Step 3

Solve for $y$ .

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Step 3.1

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

$e^{\ln (| y |)} = e^{\frac{8}{3} x^{3} - \frac{9}{2} x^{2} + C}$

Step 3.2

Rewrite $\ln (| y |) = \frac{8}{3} x^{3} - \frac{9}{2} x^{2} + C$ 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 $b^{y} = x$ .

$e^{\frac{8}{3} x^{3} - \frac{9}{2} x^{2} + C} = | y |$

Step 3.3

Solve for $y$ .

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Step 3.3.1

Rewrite the equation as $| y | = e^{\frac{8}{3} x^{3} - \frac{9}{2} x^{2} + C}$ .

$| y | = e^{\frac{8}{3} x^{3} - \frac{9}{2} x^{2} + C}$

Step 3.3.2

Simplify each term.

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Step 3.3.2.1

Combine $\frac{8}{3}$ and $x^{3}$ .

$| y | = e^{\frac{8 x^{3}}{3} - \frac{9}{2} x^{2} + C}$

Step 3.3.2.2

Combine $x^{2}$ and $\frac{9}{2}$ .

$| y | = e^{\frac{8 x^{3}}{3} - \frac{x^{2} \cdot 9}{2} + C}$

Step 3.3.2.3

Move $9$ to the left of $x^{2}$ .

$| y | = e^{\frac{8 x^{3}}{3} - \frac{9 x^{2}}{2} + C}$

Step 3.3.3

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

$y = \pm e^{\frac{8 x^{3}}{3} - \frac{9 x^{2}}{2} + C}$

Step 4

Group the constant terms together.

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Step 4.1

Rewrite $e^{\frac{8 x^{3}}{3} - \frac{9 x^{2}}{2} + C}$ as $e^{\frac{8 x^{3}}{3} - \frac{9 x^{2}}{2}} e^{C}$ .

$y = \pm e^{\frac{8 x^{3}}{3} - \frac{9 x^{2}}{2}} e^{C}$

Step 4.2

Reorder $e^{\frac{8 x^{3}}{3} - \frac{9 x^{2}}{2}}$ and $e^{C}$ .

$y = \pm e^{C} e^{\frac{8 x^{3}}{3} - \frac{9 x^{2}}{2}}$

Step 4.3

Combine constants with the plus or minus.

$y = C e^{\frac{8 x^{3}}{3} - \frac{9 x^{2}}{2}}$