Calculus Examples

$\frac{d y}{d x} = y x^{2} - 1.1 y$

Step 1

Separate the variables.

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Step 1.1

Factor.

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Step 1.1.1

Factor $y$ out of $y x^{2} - 1.1 y$ .

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Step 1.1.1.1

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

$\frac{d y}{d x} = y (x^{2}) - 1.1 y$

Step 1.1.1.2

Factor $y$ out of $- 1.1 y$ .

$\frac{d y}{d x} = y (x^{2}) + y \cdot - 1.1$

Step 1.1.1.3

Factor $y$ out of $y (x^{2}) + y \cdot - 1.1$ .

$\frac{d y}{d x} = y (x^{2} - 1.1)$

Step 1.1.2

Rewrite $1.1$ as ${1.04880884}^{2}$ .

$\frac{d y}{d x} = y (x^{2} - {1.04880884}^{2})$

Step 1.1.3

Factor.

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Step 1.1.3.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1.04880884$ .

$\frac{d y}{d x} = y ((x + 1.04880884) (x - 1.04880884))$

Step 1.1.3.2

Remove unnecessary parentheses.

$\frac{d y}{d x} = y (x + 1.04880884) (x - 1.04880884)$

Step 1.2

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

$\frac{1}{y} \frac{d y}{d x} = \frac{1}{y} (y (x + 1.04880884) (x - 1.04880884))$

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 $y (x + 1.04880884) (x - 1.04880884)$ .

$\frac{1}{y} \frac{d y}{d x} = \frac{1}{y} (y ((x + 1.04880884) (x - 1.04880884)))$

Step 1.3.1.2

Cancel the common factor.

$\frac{1}{y} \frac{d y}{d x} = \frac{1}{y} (y ((x + 1.04880884) (x - 1.04880884)))$

Step 1.3.1.3

Rewrite the expression.

$\frac{1}{y} \frac{d y}{d x} = (x + 1.04880884) (x - 1.04880884)$

Step 1.3.2

Expand $(x + 1.04880884) (x - 1.04880884)$ using the FOIL Method.

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Step 1.3.2.1

Apply the distributive property.

$\frac{1}{y} \frac{d y}{d x} = x (x - 1.04880884) + 1.04880884 (x - 1.04880884)$

Step 1.3.2.2

Apply the distributive property.

$\frac{1}{y} \frac{d y}{d x} = x \cdot x + x \cdot - 1.04880884 + 1.04880884 (x - 1.04880884)$

Step 1.3.2.3

Apply the distributive property.

$\frac{1}{y} \frac{d y}{d x} = x \cdot x + x \cdot - 1.04880884 + 1.04880884 x + 1.04880884 \cdot - 1.04880884$

Step 1.3.3

Combine the opposite terms in $x \cdot x + x \cdot - 1.04880884 + 1.04880884 x + 1.04880884 \cdot - 1.04880884$ .

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Step 1.3.3.1

Reorder the factors in the terms $x \cdot - 1.04880884$ and $1.04880884 x$ .

$\frac{1}{y} \frac{d y}{d x} = x \cdot x - 1.04880884 x + 1.04880884 x + 1.04880884 \cdot - 1.04880884$

Step 1.3.3.2

Add $- 1.04880884 x$ and $1.04880884 x$ .

$\frac{1}{y} \frac{d y}{d x} = x \cdot x + 0 x + 1.04880884 \cdot - 1.04880884$

Step 1.3.4

Simplify each term.

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Step 1.3.4.1

Multiply $x$ by $x$ .

$\frac{1}{y} \frac{d y}{d x} = x^{2} + 0 x + 1.04880884 \cdot - 1.04880884$

Step 1.3.4.2

Multiply $0$ by $x$ .

$\frac{1}{y} \frac{d y}{d x} = x^{2} + 0 + 1.04880884 \cdot - 1.04880884$

Step 1.3.4.3

Multiply $1.04880884$ by $- 1.04880884$ .

$\frac{1}{y} \frac{d y}{d x} = x^{2} + 0 - 1.1$

Step 1.3.5

Add $x^{2}$ and $0$ .

$\frac{1}{y} \frac{d y}{d x} = x^{2} - 1.1$

Step 1.4

Rewrite the equation.

$\frac{1}{y} d y = (x^{2} - 1.1) d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1}{y} d y = \int x^{2} - 1.1 d x$

Step 2.2

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

$\ln (| y |) + C_{1} = \int x^{2} - 1.1 d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

$\ln (| y |) + C_{1} = \int x^{2} d x + \int - 1.1 d x$

Step 2.3.2

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

$\ln (| y |) + C_{1} = \frac{1}{3} x^{3} + C_{2} + \int - 1.1 d x$

Step 2.3.3

Apply the constant rule.

$\ln (| y |) + C_{1} = \frac{1}{3} x^{3} + C_{2} - 1.1 x + C_{3}$

Step 2.3.4

Simplify.

$\ln (| y |) + C_{1} = \frac{1}{3} x^{3} - 1.1 x + C_{4}$

Step 2.4

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

$\ln (| y |) = \frac{1}{3} x^{3} - 1.1 x + C$

Step 3

Solve for $y$ .

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Step 3.1

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

$e^{\ln (| y |)} = e^{\frac{1}{3} x^{3} - 1.1 x + C}$

Step 3.2

Rewrite $\ln (| y |) = \frac{1}{3} x^{3} - 1.1 x + C$ 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 $b^{y} = x$ .

$e^{\frac{1}{3} x^{3} - 1.1 x + C} = | y |$

Step 3.3

Solve for $y$ .

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Step 3.3.1

Rewrite the equation as $| y | = e^{\frac{1}{3} x^{3} - 1.1 x + C}$ .

$| y | = e^{\frac{1}{3} x^{3} - 1.1 x + C}$

Step 3.3.2

Combine $\frac{1}{3}$ and $x^{3}$ .

$| y | = e^{\frac{x^{3}}{3} - 1.1 x + C}$

Step 3.3.3

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

$y = \pm e^{\frac{x^{3}}{3} - 1.1 x + C}$

Step 4

Group the constant terms together.

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Step 4.1

Rewrite $e^{\frac{x^{3}}{3} - 1.1 x + C}$ as $e^{\frac{x^{3}}{3} - 1.1 x} e^{C}$ .

$y = \pm e^{\frac{x^{3}}{3} - 1.1 x} e^{C}$

Step 4.2

Reorder $e^{\frac{x^{3}}{3} - 1.1 x}$ and $e^{C}$ .

$y = \pm e^{C} e^{\frac{x^{3}}{3} - 1.1 x}$

Step 4.3

Combine constants with the plus or minus.

$y = C e^{\frac{x^{3}}{3} - 1.1 x}$