Calculus Examples

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

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 $y t^{2} - 1.1 y$ .

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Step 1.1.1.1

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

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

Step 1.1.1.2

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

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

Step 1.1.1.3

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

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

Step 1.1.2

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

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

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 $a = t$ and $b = 1.04880884$ .

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

Step 1.1.3.2

Remove unnecessary parentheses.

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

Step 1.2

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

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

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

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

Step 1.3.1.2

Cancel the common factor.

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

Step 1.3.1.3

Rewrite the expression.

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

Step 1.3.2

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

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Step 1.3.2.1

Apply the distributive property.

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

Step 1.3.2.2

Apply the distributive property.

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

Step 1.3.2.3

Apply the distributive property.

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

Step 1.3.3

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

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Step 1.3.3.1

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

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

Step 1.3.3.2

Add $- 1.04880884 t$ and $1.04880884 t$ .

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

Step 1.3.4

Simplify each term.

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Step 1.3.4.1

Multiply $t$ by $t$ .

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

Step 1.3.4.2

Multiply $0$ by $t$ .

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

Step 1.3.4.3

Multiply $1.04880884$ by $- 1.04880884$ .

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

Step 1.3.5

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

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

Step 1.4

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

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

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

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

Step 2.3.2

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

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

Step 2.3.3

Apply the constant rule.

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

Step 2.3.4

Simplify.

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

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

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

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

Step 3.2

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

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

Step 3.3

Solve for $y$ .

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Step 3.3.1

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

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

Step 3.3.2

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

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

Step 3.3.3

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

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

Step 4

Group the constant terms together.

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Step 4.1

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

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

Step 4.2

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

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

Step 4.3

Combine constants with the plus or minus.

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