Calculus Examples

$\frac{d Q}{d t} + \frac{2}{10 + 2 t} Q = 4$

Step 1

The integrating factor is defined by the formula $e^{\int P (t) d t}$ , where $P (t) = \frac{2}{10 + 2 t}$ .

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Step 1.1

Set up the integration.

$e^{\int \frac{2}{10 + 2 t} d t}$

Step 1.2

Integrate $\frac{2}{10 + 2 t}$ .

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Step 1.2.1

Cancel the common factor of $2$ and $10 + 2 t$ .

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Step 1.2.1.1

Factor $2$ out of $2$ .

$e^{\int \frac{2 \cdot 1}{10 + 2 t} d t}$

Step 1.2.1.2

Cancel the common factors.

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Step 1.2.1.2.1

Factor $2$ out of $10$ .

$e^{\int \frac{2 \cdot 1}{2 (5) + 2 t} d t}$

Step 1.2.1.2.2

Factor $2$ out of $2 t$ .

$e^{\int \frac{2 \cdot 1}{2 (5) + 2 (t)} d t}$

Step 1.2.1.2.3

Factor $2$ out of $2 (5) + 2 (t)$ .

$e^{\int \frac{2 \cdot 1}{2 (5 + t)} d t}$

Step 1.2.1.2.4

Cancel the common factor.

$e^{\int \frac{2 \cdot 1}{2 (5 + t)} d t}$

Step 1.2.1.2.5

Rewrite the expression.

$e^{\int \frac{1}{5 + t} d t}$

Step 1.2.2

Let $u = 5 + t$ . Then $d u = d t$ . Rewrite using $u$ and $d$ $u$ .

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Step 1.2.2.1

Let $u = 5 + t$ . Find $\frac{d u}{d t}$ .

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Step 1.2.2.1.1

Differentiate $5 + t$ .

$\frac{d}{d t} [5 + t]$

Step 1.2.2.1.2

By the Sum Rule, the derivative of $5 + t$ with respect to $t$ is $\frac{d}{d t} [5] + \frac{d}{d t} [t]$ .

$\frac{d}{d t} [5] + \frac{d}{d t} [t]$

Step 1.2.2.1.3

Since $5$ is constant with respect to $t$ , the derivative of $5$ with respect to $t$ is $0$ .

$0 + \frac{d}{d t} [t]$

Step 1.2.2.1.4

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 1$ .

$0 + 1$

Step 1.2.2.1.5

Add $0$ and $1$ .

$1$

Step 1.2.2.2

Rewrite the problem using $u$ and $d u$ .

$e^{\int \frac{1}{u} d u}$

Step 1.2.3

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

$e^{\ln (| u |) + C}$

Step 1.2.4

Replace all occurrences of $u$ with $5 + t$ .

$e^{\ln (| 5 + t |) + C}$

Step 1.3

Remove the constant of integration.

$e^{\ln (5 + t)}$

Step 1.4

Exponentiation and log are inverse functions.

$5 + t$

Step 2

Multiply each term by the integrating factor $5 + t$ .

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Step 2.1

Multiply each term by $5 + t$ .

$(5 + t) \frac{d Q}{d t} + (5 + t) (\frac{2}{10 + 2 t} Q) = (5 + t) \cdot 4$

Step 2.2

Simplify each term.

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Step 2.2.1

Apply the distributive property.

$5 \frac{d Q}{d t} + t \frac{d Q}{d t} + (5 + t) (\frac{2}{10 + 2 t} Q) = (5 + t) \cdot 4$

Step 2.2.2

Factor $2$ out of $10 + 2 t$ .

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Step 2.2.2.1

Factor $2$ out of $10$ .

$5 \frac{d Q}{d t} + t \frac{d Q}{d t} + (5 + t) (\frac{2}{2 \cdot 5 + 2 t} Q) = (5 + t) \cdot 4$

Step 2.2.2.2

Factor $2$ out of $2 \cdot 5 + 2 t$ .

$5 \frac{d Q}{d t} + t \frac{d Q}{d t} + (5 + t) (\frac{2}{2 (5 + t)} Q) = (5 + t) \cdot 4$

Step 2.2.3

Cancel the common factor of $2$ .

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Step 2.2.3.1

Cancel the common factor.

$5 \frac{d Q}{d t} + t \frac{d Q}{d t} + (5 + t) (\frac{2}{2 (5 + t)} Q) = (5 + t) \cdot 4$

Step 2.2.3.2

Rewrite the expression.

$5 \frac{d Q}{d t} + t \frac{d Q}{d t} + (5 + t) (\frac{1}{5 + t} Q) = (5 + t) \cdot 4$

Step 2.2.4

Combine $\frac{1}{5 + t}$ and $Q$ .

$5 \frac{d Q}{d t} + t \frac{d Q}{d t} + (5 + t) \frac{Q}{5 + t} = (5 + t) \cdot 4$

Step 2.2.5

Cancel the common factor of $5 + t$ .

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Step 2.2.5.1

Cancel the common factor.

$5 \frac{d Q}{d t} + t \frac{d Q}{d t} + (5 + t) \frac{Q}{5 + t} = (5 + t) \cdot 4$

Step 2.2.5.2

Rewrite the expression.

$5 \frac{d Q}{d t} + t \frac{d Q}{d t} + Q = (5 + t) \cdot 4$

Step 2.3

Apply the distributive property.

$5 \frac{d Q}{d t} + t \frac{d Q}{d t} + Q = 5 \cdot 4 + t \cdot 4$

Step 2.4

Multiply $5$ by $4$ .

$5 \frac{d Q}{d t} + t \frac{d Q}{d t} + Q = 20 + t \cdot 4$

Step 2.5

Move $4$ to the left of $t$ .

$5 \frac{d Q}{d t} + t \frac{d Q}{d t} + Q = 20 + 4 t$

Step 3

Rewrite the left side as a result of differentiating a product.

$\frac{d}{d t} [(5 + t) Q] = 20 + 4 t$

Step 4

Set up an integral on each side.

$\int \frac{d}{d t} [(5 + t) Q] d t = \int 20 + 4 t d t$

Step 5

Integrate the left side.

$(5 + t) Q = \int 20 + 4 t d t$

Step 6

Integrate the right side.

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Step 6.1

Split the single integral into multiple integrals.

$(5 + t) Q = \int 20 d t + \int 4 t d t$

Step 6.2

Apply the constant rule.

$(5 + t) Q = 20 t + C + \int 4 t d t$

Step 6.3

Since $4$ is constant with respect to $t$ , move $4$ out of the integral.

$(5 + t) Q = 20 t + C + 4 \int t d t$

Step 6.4

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

$(5 + t) Q = 20 t + C + 4 (\frac{1}{2} t^{2} + C)$

Step 6.5

Simplify.

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Step 6.5.1

Simplify.

$(5 + t) Q = 20 t + 4 (\frac{1}{2}) t^{2} + C$

Step 6.5.2

Simplify.

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Step 6.5.2.1

Combine $4$ and $\frac{1}{2}$ .

$(5 + t) Q = 20 t + \frac{4}{2} t^{2} + C$

Step 6.5.2.2

Cancel the common factor of $4$ and $2$ .

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Step 6.5.2.2.1

Factor $2$ out of $4$ .

$(5 + t) Q = 20 t + \frac{2 \cdot 2}{2} t^{2} + C$

Step 6.5.2.2.2

Cancel the common factors.

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Step 6.5.2.2.2.1

Factor $2$ out of $2$ .

$(5 + t) Q = 20 t + \frac{2 \cdot 2}{2 (1)} t^{2} + C$

Step 6.5.2.2.2.2

Cancel the common factor.

$(5 + t) Q = 20 t + \frac{2 \cdot 2}{2 \cdot 1} t^{2} + C$

Step 6.5.2.2.2.3

Rewrite the expression.

$(5 + t) Q = 20 t + \frac{2}{1} t^{2} + C$

Step 6.5.2.2.2.4

Divide $2$ by $1$ .

$(5 + t) Q = 20 t + 2 t^{2} + C$

Step 7

Divide each term in $(5 + t) Q = 20 t + 2 t^{2} + C$ by $5 + t$ and simplify.

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Step 7.1

Divide each term in $(5 + t) Q = 20 t + 2 t^{2} + C$ by $5 + t$ .

$\frac{(5 + t) Q}{5 + t} = \frac{20 t}{5 + t} + \frac{2 t^{2}}{5 + t} + \frac{C}{5 + t}$

Step 7.2

Simplify the left side.

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Step 7.2.1

Cancel the common factor of $5 + t$ .

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Step 7.2.1.1

Cancel the common factor.

$\frac{(5 + t) Q}{5 + t} = \frac{20 t}{5 + t} + \frac{2 t^{2}}{5 + t} + \frac{C}{5 + t}$

Step 7.2.1.2

Divide $Q$ by $1$ .

$Q = \frac{20 t}{5 + t} + \frac{2 t^{2}}{5 + t} + \frac{C}{5 + t}$

Step 7.3

Simplify the right side.

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Step 7.3.1

Combine the numerators over the common denominator.

$Q = \frac{20 t + 2 t^{2}}{5 + t} + \frac{C}{5 + t}$

Step 7.3.2

Combine the numerators over the common denominator.

$Q = \frac{20 t + 2 t^{2} + C}{5 + t}$