Calculus Examples

$\frac{d y}{d t} = t + 2 e^{t - 3}$ ; and , $y (3) = 4$

Step 1

Rewrite the equation.

$d y = (t + 2 e^{t - 3}) d t$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d y = \int t + 2 e^{t - 3} d t$

Step 2.2

Apply the constant rule.

$y + C_{1} = \int t + 2 e^{t - 3} d t$

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

$y + C_{1} = \int t d t + \int 2 e^{t - 3} d t$

Step 2.3.2

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

$y + C_{1} = \frac{1}{2} t^{2} + C_{2} + \int 2 e^{t - 3} d t$

Step 2.3.3

Since $2$ is constant with respect to $t$ , move $2$ out of the integral.

$y + C_{1} = \frac{1}{2} t^{2} + C_{2} + 2 \int e^{t - 3} d t$

Step 2.3.4

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

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Step 2.3.4.1

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

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Step 2.3.4.1.1

Differentiate $t - 3$ .

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

Step 2.3.4.1.2

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

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

Step 2.3.4.1.3

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

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

Step 2.3.4.1.4

Since $- 3$ is constant with respect to $t$ , the derivative of $- 3$ with respect to $t$ is $0$ .

$1 + 0$

Step 2.3.4.1.5

Add $1$ and $0$ .

$1$

Step 2.3.4.2

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

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

Step 2.3.5

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

$y + C_{1} = \frac{1}{2} t^{2} + C_{2} + 2 (e^{u} + C_{3})$

Step 2.3.6

Simplify.

$y + C_{1} = \frac{1}{2} t^{2} + 2 e^{u} + C_{4}$

Step 2.3.7

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

$y + C_{1} = \frac{1}{2} t^{2} + 2 e^{t - 3} + C_{4}$

Step 2.4

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

$y = \frac{1}{2} t^{2} + 2 e^{t - 3} + K$

Step 3

Use the initial condition to find the value of $K$ by substituting $3$ for $t$ and $4$ for $y$ in $y = \frac{1}{2} t^{2} + 2 e^{t - 3} + K$ .

$4 = \frac{1}{2} \cdot 3^{2} + 2 e^{3 - 3} + K$

Step 4

Solve for $K$ .

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Step 4.1

Rewrite the equation as $\frac{1}{2} \cdot 3^{2} + 2 e^{3 - 3} + K = 4$ .

$\frac{1}{2} \cdot 3^{2} + 2 e^{3 - 3} + K = 4$

Step 4.2

Simplify $\frac{1}{2} \cdot 3^{2} + 2 e^{3 - 3} + K$ .

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Step 4.2.1

Subtract $3$ from $3$ .

$\frac{1}{2} \cdot 3^{2} + 2 e^{0} + K = 4$

Step 4.2.2

Simplify each term.

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Step 4.2.2.1

Raise $3$ to the power of $2$ .

$\frac{1}{2} \cdot 9 + 2 e^{0} + K = 4$

Step 4.2.2.2

Combine $\frac{1}{2}$ and $9$ .

$\frac{9}{2} + 2 e^{0} + K = 4$

Step 4.2.2.3

Anything raised to $0$ is $1$ .

$\frac{9}{2} + 2 \cdot 1 + K = 4$

Step 4.2.2.4

Multiply $2$ by $1$ .

$\frac{9}{2} + 2 + K = 4$

Step 4.2.3

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

$\frac{9}{2} + 2 \cdot \frac{2}{2} + K = 4$

Step 4.2.4

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

$\frac{9}{2} + \frac{2 \cdot 2}{2} + K = 4$

Step 4.2.5

Combine the numerators over the common denominator.

$\frac{9 + 2 \cdot 2}{2} + K = 4$

Step 4.2.6

Simplify the numerator.

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Step 4.2.6.1

Multiply $2$ by $2$ .

$\frac{9 + 4}{2} + K = 4$

Step 4.2.6.2

Add $9$ and $4$ .

$\frac{13}{2} + K = 4$

Step 4.3

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

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Step 4.3.1

Subtract $\frac{13}{2}$ from both sides of the equation.

$K = 4 - \frac{13}{2}$

Step 4.3.2

To write $4$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$K = 4 \cdot \frac{2}{2} - \frac{13}{2}$

Step 4.3.3

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

$K = \frac{4 \cdot 2}{2} - \frac{13}{2}$

Step 4.3.4

Combine the numerators over the common denominator.

$K = \frac{4 \cdot 2 - 13}{2}$

Step 4.3.5

Simplify the numerator.

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Step 4.3.5.1

Multiply $4$ by $2$ .

$K = \frac{8 - 13}{2}$

Step 4.3.5.2

Subtract $13$ from $8$ .

$K = \frac{- 5}{2}$

Step 4.3.6

Move the negative in front of the fraction.

$K = - \frac{5}{2}$

Step 5

Substitute $- \frac{5}{2}$ for $K$ in $y = \frac{1}{2} t^{2} + 2 e^{t - 3} + K$ and simplify.

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Step 5.1

Substitute $- \frac{5}{2}$ for $K$ .

$y = \frac{1}{2} t^{2} + 2 e^{t - 3} - \frac{5}{2}$

Step 5.2

Combine $\frac{1}{2}$ and $t^{2}$ .

$y = \frac{t^{2}}{2} + 2 e^{t - 3} - \frac{5}{2}$

Step 5.3

To write $2 e^{t - 3}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$y = \frac{t^{2}}{2} + 2 e^{t - 3} \cdot \frac{2}{2} - \frac{5}{2}$

Step 5.4

Combine $2 e^{t - 3}$ and $\frac{2}{2}$ .

$y = \frac{t^{2}}{2} + \frac{2 e^{t - 3} \cdot 2}{2} - \frac{5}{2}$

Step 5.5

Combine the numerators over the common denominator.

$y = \frac{t^{2} + 2 e^{t - 3} \cdot 2}{2} - \frac{5}{2}$

Step 5.6

Multiply $2$ by $2$ .

$y = \frac{t^{2} + 4 e^{t - 3}}{2} - \frac{5}{2}$