Calculus Examples

$\frac{d y}{d t} + 2 y = 1$ , $y (0) = \frac{2}{5}$

Step 1

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

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Step 1.1

Set up the integration.

$e^{\int 2 d t}$

Step 1.2

Apply the constant rule.

$e^{2 t + C}$

Step 1.3

Remove the constant of integration.

$e^{2 t}$

Step 2

Multiply each term by the integrating factor $e^{2 t}$ .

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Step 2.1

Multiply each term by $e^{2 t}$ .

$e^{2 t} \frac{d y}{d t} + e^{2 t} (2 y) = e^{2 t} \cdot 1$

Step 2.2

Rewrite using the commutative property of multiplication.

$e^{2 t} \frac{d y}{d t} + 2 e^{2 t} y = e^{2 t} \cdot 1$

Step 2.3

Multiply $e^{2 t}$ by $1$ .

$e^{2 t} \frac{d y}{d t} + 2 e^{2 t} y = e^{2 t}$

Step 2.4

Reorder factors in $e^{2 t} \frac{d y}{d t} + 2 e^{2 t} y = e^{2 t}$ .

$e^{2 t} \frac{d y}{d t} + 2 y e^{2 t} = e^{2 t}$

Step 3

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

$\frac{d}{d t} [e^{2 t} y] = e^{2 t}$

Step 4

Set up an integral on each side.

$\int \frac{d}{d t} [e^{2 t} y] d t = \int e^{2 t} d t$

Step 5

Integrate the left side.

$e^{2 t} y = \int e^{2 t} d t$

Step 6

Integrate the right side.

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Step 6.1

Let $u = 2 t$ . Then $d u = 2 d t$ , so $\frac{1}{2} d u = d t$ . Rewrite using $u$ and $d$ $u$ .

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Step 6.1.1

Let $u = 2 t$ . Find $\frac{d u}{d t}$ .

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Step 6.1.1.1

Differentiate $2 t$ .

$\frac{d}{d t} [2 t]$

Step 6.1.1.2

Since $2$ is constant with respect to $t$ , the derivative of $2 t$ with respect to $t$ is $2 \frac{d}{d t} [t]$ .

$2 \frac{d}{d t} [t]$

Step 6.1.1.3

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

$2 \cdot 1$

Step 6.1.1.4

Multiply $2$ by $1$ .

$2$

Step 6.1.2

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

$e^{2 t} y = \int e^{u} \frac{1}{2} d u$

Step 6.2

Combine $e^{u}$ and $\frac{1}{2}$ .

$e^{2 t} y = \int \frac{e^{u}}{2} d u$

Step 6.3

Since $\frac{1}{2}$ is constant with respect to $u$ , move $\frac{1}{2}$ out of the integral.

$e^{2 t} y = \frac{1}{2} \int e^{u} d u$

Step 6.4

The integral of $e^{u}$ with respect to $u$ is $e^{u}$ .

$e^{2 t} y = \frac{1}{2} (e^{u} + C)$

Step 6.5

Simplify.

$e^{2 t} y = \frac{1}{2} e^{u} + C$

Step 6.6

Replace all occurrences of $u$ with $2 t$ .

$e^{2 t} y = \frac{1}{2} e^{2 t} + C$

Step 7

Divide each term in $e^{2 t} y = \frac{1}{2} e^{2 t} + C$ by $e^{2 t}$ and simplify.

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Step 7.1

Divide each term in $e^{2 t} y = \frac{1}{2} e^{2 t} + C$ by $e^{2 t}$ .

$\frac{e^{2 t} y}{e^{2 t}} = \frac{\frac{1}{2} e^{2 t}}{e^{2 t}} + \frac{C}{e^{2 t}}$

Step 7.2

Simplify the left side.

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Step 7.2.1

Cancel the common factor of $e^{2 t}$ .

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Step 7.2.1.1

Cancel the common factor.

$\frac{e^{2 t} y}{e^{2 t}} = \frac{\frac{1}{2} e^{2 t}}{e^{2 t}} + \frac{C}{e^{2 t}}$

Step 7.2.1.2

Divide $y$ by $1$ .

$y = \frac{\frac{1}{2} e^{2 t}}{e^{2 t}} + \frac{C}{e^{2 t}}$

Step 7.3

Simplify the right side.

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Step 7.3.1

Cancel the common factor of $e^{2 t}$ .

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Step 7.3.1.1

Cancel the common factor.

$y = \frac{\frac{1}{2} e^{2 t}}{e^{2 t}} + \frac{C}{e^{2 t}}$

Step 7.3.1.2

Divide $\frac{1}{2}$ by $1$ .

$y = \frac{1}{2} + \frac{C}{e^{2 t}}$

Step 8

Use the initial condition to find the value of $C$ by substituting $0$ for $t$ and $\frac{2}{5}$ for $y$ in $y = \frac{1}{2} + \frac{C}{e^{2 t}}$ .

$\frac{2}{5} = \frac{1}{2} + \frac{C}{e^{2 \cdot 0}}$

Step 9

Solve for $C$ .

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Step 9.1

Rewrite the equation as $\frac{1}{2} + \frac{C}{e^{2 \cdot 0}} = \frac{2}{5}$ .

$\frac{1}{2} + \frac{C}{e^{2 \cdot 0}} = \frac{2}{5}$

Step 9.2

Simplify each term.

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Step 9.2.1

Simplify the denominator.

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Step 9.2.1.1

Multiply $2$ by $0$ .

$\frac{1}{2} + \frac{C}{e^{0}} = \frac{2}{5}$

Step 9.2.1.2

Anything raised to $0$ is $1$ .

$\frac{1}{2} + \frac{C}{1} = \frac{2}{5}$

Step 9.2.2

Divide $C$ by $1$ .

$\frac{1}{2} + C = \frac{2}{5}$

Step 9.3

Move all terms not containing $C$ to the right side of the equation.

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Step 9.3.1

Subtract $\frac{1}{2}$ from both sides of the equation.

$C = \frac{2}{5} - \frac{1}{2}$

Step 9.3.2

To write $\frac{2}{5}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$C = \frac{2}{5} \cdot \frac{2}{2} - \frac{1}{2}$

Step 9.3.3

To write $- \frac{1}{2}$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$C = \frac{2}{5} \cdot \frac{2}{2} - \frac{1}{2} \cdot \frac{5}{5}$

Step 9.3.4

Write each expression with a common denominator of $10$ , by multiplying each by an appropriate factor of $1$ .

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Step 9.3.4.1

Multiply $\frac{2}{5}$ by $\frac{2}{2}$ .

$C = \frac{2 \cdot 2}{5 \cdot 2} - \frac{1}{2} \cdot \frac{5}{5}$

Step 9.3.4.2

Multiply $5$ by $2$ .

$C = \frac{2 \cdot 2}{10} - \frac{1}{2} \cdot \frac{5}{5}$

Step 9.3.4.3

Multiply $\frac{1}{2}$ by $\frac{5}{5}$ .

$C = \frac{2 \cdot 2}{10} - \frac{5}{2 \cdot 5}$

Step 9.3.4.4

Multiply $2$ by $5$ .

$C = \frac{2 \cdot 2}{10} - \frac{5}{10}$

Step 9.3.5

Combine the numerators over the common denominator.

$C = \frac{2 \cdot 2 - 5}{10}$

Step 9.3.6

Simplify the numerator.

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Step 9.3.6.1

Multiply $2$ by $2$ .

$C = \frac{4 - 5}{10}$

Step 9.3.6.2

Subtract $5$ from $4$ .

$C = \frac{- 1}{10}$

Step 9.3.7

Move the negative in front of the fraction.

$C = - \frac{1}{10}$

Step 10

Substitute $- \frac{1}{10}$ for $C$ in $y = \frac{1}{2} + \frac{C}{e^{2 t}}$ and simplify.

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Step 10.1

Substitute $- \frac{1}{10}$ for $C$ .

$y = \frac{1}{2} + \frac{- \frac{1}{10}}{e^{2 t}}$

Step 10.2

Simplify each term.

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Step 10.2.1

Multiply the numerator by the reciprocal of the denominator.

$y = \frac{1}{2} - \frac{1}{10} \cdot \frac{1}{e^{2 t}}$

Step 10.2.2

Multiply $\frac{1}{e^{2 t}}$ by $\frac{1}{10}$ .

$y = \frac{1}{2} - \frac{1}{e^{2 t} \cdot 10}$

Step 10.2.3

Move $10$ to the left of $e^{2 t}$ .

$y = \frac{1}{2} - \frac{1}{10 e^{2 t}}$