Calculus Examples

$\frac{t}{5} \cdot \frac{d y}{d t} = (y - 3)$

Step 1

Separate the variables.

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Step 1.1

Solve for $\frac{d y}{d t}$ .

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Step 1.1.1

Combine $\frac{t}{5}$ and $\frac{d y}{d t}$ .

$\frac{t \frac{d y}{d t}}{5} = y - 3$

Step 1.1.2

Multiply both sides of the equation by $5$ .

$5 \frac{t \frac{d y}{d t}}{5} = 5 (y - 3)$

Step 1.1.3

Simplify both sides of the equation.

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Step 1.1.3.1

Simplify the left side.

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Step 1.1.3.1.1

Cancel the common factor of $5$ .

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Step 1.1.3.1.1.1

Cancel the common factor.

$5 \frac{t \frac{d y}{d t}}{5} = 5 (y - 3)$

Step 1.1.3.1.1.2

Rewrite the expression.

$t \frac{d y}{d t} = 5 (y - 3)$

Step 1.1.3.2

Simplify the right side.

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Step 1.1.3.2.1

Simplify $5 (y - 3)$ .

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Step 1.1.3.2.1.1

Apply the distributive property.

$t \frac{d y}{d t} = 5 y + 5 \cdot - 3$

Step 1.1.3.2.1.2

Multiply $5$ by $- 3$ .

$t \frac{d y}{d t} = 5 y - 15$

Step 1.1.4

Divide each term in $t \frac{d y}{d t} = 5 y - 15$ by $t$ and simplify.

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Step 1.1.4.1

Divide each term in $t \frac{d y}{d t} = 5 y - 15$ by $t$ .

$\frac{t \frac{d y}{d t}}{t} = \frac{5 y}{t} + \frac{- 15}{t}$

Step 1.1.4.2

Simplify the left side.

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Step 1.1.4.2.1

Cancel the common factor of $t$ .

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Step 1.1.4.2.1.1

Cancel the common factor.

$\frac{t \frac{d y}{d t}}{t} = \frac{5 y}{t} + \frac{- 15}{t}$

Step 1.1.4.2.1.2

Divide $\frac{d y}{d t}$ by $1$ .

$\frac{d y}{d t} = \frac{5 y}{t} + \frac{- 15}{t}$

Step 1.1.4.3

Simplify the right side.

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Step 1.1.4.3.1

Move the negative in front of the fraction.

$\frac{d y}{d t} = \frac{5 y}{t} - \frac{15}{t}$

Step 1.2

Factor.

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Step 1.2.1

Combine the numerators over the common denominator.

$\frac{d y}{d t} = \frac{5 y - 15}{t}$

Step 1.2.2

Factor $5$ out of $5 y - 15$ .

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Step 1.2.2.1

Factor $5$ out of $5 y$ .

$\frac{d y}{d t} = \frac{5 (y) - 15}{t}$

Step 1.2.2.2

Factor $5$ out of $- 15$ .

$\frac{d y}{d t} = \frac{5 y + 5 \cdot - 3}{t}$

Step 1.2.2.3

Factor $5$ out of $5 y + 5 \cdot - 3$ .

$\frac{d y}{d t} = \frac{5 (y - 3)}{t}$

Step 1.3

Multiply both sides by $\frac{1}{y - 3}$ .

$\frac{1}{y - 3} \frac{d y}{d t} = \frac{1}{y - 3} \cdot \frac{5 (y - 3)}{t}$

Step 1.4

Cancel the common factor of $y - 3$ .

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Step 1.4.1

Factor $y - 3$ out of $5 (y - 3)$ .

$\frac{1}{y - 3} \frac{d y}{d t} = \frac{1}{y - 3} \cdot \frac{(y - 3) \cdot 5}{t}$

Step 1.4.2

Cancel the common factor.

$\frac{1}{y - 3} \frac{d y}{d t} = \frac{1}{y - 3} \cdot \frac{(y - 3) \cdot 5}{t}$

Step 1.4.3

Rewrite the expression.

$\frac{1}{y - 3} \frac{d y}{d t} = \frac{5}{t}$

Step 1.5

Rewrite the equation.

$\frac{1}{y - 3} d y = \frac{5}{t} d t$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1}{y - 3} d y = \int \frac{5}{t} d t$

Step 2.2

Integrate the left side.

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Step 2.2.1

Let $u = y - 3$ . Then $d u = d y$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.2.1.1

Let $u = y - 3$ . Find $\frac{d u}{d y}$ .

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Step 2.2.1.1.1

Differentiate $y - 3$ .

$\frac{d}{d y} [y - 3]$

Step 2.2.1.1.2

By the Sum Rule, the derivative of $y - 3$ with respect to $y$ is $\frac{d}{d y} [y] + \frac{d}{d y} [- 3]$ .

$\frac{d}{d y} [y] + \frac{d}{d y} [- 3]$

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 $n = 1$ .

$1 + \frac{d}{d y} [- 3]$

Step 2.2.1.1.4

Since $- 3$ is constant with respect to $y$ , the derivative of $- 3$ 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 $d u$ .

$\int \frac{1}{u} d u = \int \frac{5}{t} d t$

Step 2.2.2

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

$\ln (| u |) + C_{1} = \int \frac{5}{t} d t$

Step 2.2.3

Replace all occurrences of $u$ with $y - 3$ .

$\ln (| y - 3 |) + C_{1} = \int \frac{5}{t} d t$

Step 2.3

Integrate the right side.

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Step 2.3.1

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

$\ln (| y - 3 |) + C_{1} = 5 \int \frac{1}{t} d t$

Step 2.3.2

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

$\ln (| y - 3 |) + C_{1} = 5 (\ln (| t |) + C_{2})$

Step 2.3.3

Simplify.

$\ln (| y - 3 |) + C_{1} = 5 \ln (| t |) + C_{2}$

Step 2.4

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

$\ln (| y - 3 |) = 5 \ln (| t |) + C$

Step 3

Solve for $y$ .

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Step 3.1

Move all the terms containing a logarithm to the left side of the equation.

$\ln (| y - 3 |) - 5 \ln (| t |) = C$

Step 3.2

Simplify the left side.

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Step 3.2.1

Simplify $\ln (| y - 3 |) - 5 \ln (| t |)$ .

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Step 3.2.1.1

Simplify $- 5 \ln (| t |)$ by moving $5$ inside the logarithm.

$\ln (| y - 3 |) - \ln ({| t |}^{5}) = C$

Step 3.2.1.2

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

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

Step 3.3

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

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

Step 3.4

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

$e^{C} = \frac{| y - 3 |}{{| t |}^{5}}$

Step 3.5

Solve for $y$ .

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Step 3.5.1

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

$\frac{| y - 3 |}{{| t |}^{5}} = e^{C}$

Step 3.5.2

Multiply both sides by ${| t |}^{5}$ .

$\frac{| y - 3 |}{{| t |}^{5}} {| t |}^{5} = e^{C} {| t |}^{5}$

Step 3.5.3

Simplify the left side.

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Step 3.5.3.1

Cancel the common factor of ${| t |}^{5}$ .

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Step 3.5.3.1.1

Cancel the common factor.

$\frac{| y - 3 |}{{| t |}^{5}} {| t |}^{5} = e^{C} {| t |}^{5}$

Step 3.5.3.1.2

Rewrite the expression.

$| y - 3 | = e^{C} {| t |}^{5}$

Step 3.5.4

Solve for $y$ .

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Step 3.5.4.1

Reorder factors in $e^{C} {| t |}^{5}$ .

$| y - 3 | = {| t |}^{5} e^{C}$

Step 3.5.4.2

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

$y - 3 = \pm {| t |}^{5} e^{C}$

Step 3.5.4.3

Reorder factors in $\pm {| t |}^{5} e^{C}$ .

$y - 3 = \pm e^{C} {| t |}^{5}$

Step 3.5.4.4

Add $3$ to both sides of the equation.

$y = \pm e^{C} {| t |}^{5} + 3$

Step 4

Group the constant terms together.

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Step 4.1

Simplify the constant of integration.

$y = \pm C {| t |}^{5} + 3$

Step 4.2

Combine constants with the plus or minus.

$y = C {| t |}^{5} + 3$