Calculus Examples

$\frac{d y}{d t} = e^{t}$ , $y (0) = 1$ , $t = 5$ , $h = 0.5$

Step 1

Define $f (t, y)$ such that $\frac{d y}{d t} = f (t, y)$ .

$f (t, y) = e^{t}$

Step 2

Find $f (0, 1)$ .

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Step 2.1

Substitute $0$ for $t$ and $1$ for $y$ .

$f (0, 1) = e^{0}$

Step 2.2

Simplify.

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Step 2.2.1

Replace $e$ with an approximation.

$f (0, 1) = {2.71828182}^{0}$

Step 2.2.2

Raise $2.71828182$ to the power of $0$ .

$f (0, 1) = 1$

Step 3

Use the recursive formula $y_{1} = y_{0} + h f (t_{0}, y_{0})$ to find $y_{1}$ .

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Step 3.1

Substitute.

$y_{1} = 1 + 0.5 \cdot 1$

Step 3.2

Simplify.

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Step 3.2.1

Multiply $0.5$ by $1$ .

$y_{1} = 1 + 0.5$

Step 3.2.2

Add $1$ and $0.5$ .

$y_{1} = 1.5$

Step 4

Use the recursive formula $t_{1} = t_{0} + h$ to find $t_{1}$ .

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Step 4.1

Substitute.

$t_{1} = 0 + 0.5$

Step 4.2

Add $0$ and $0.5$ .

$t_{1} = 0.5$

Step 5

Find $f (0.5, 1.5)$ .

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Step 5.1

Substitute $0.5$ for $t$ and $1.5$ for $y$ .

$f (0.5, 1.5) = e^{0.5}$

Step 5.2

Simplify.

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Step 5.2.1

Replace $e$ with an approximation.

$f (0.5, 1.5) = {2.71828182}^{0.5}$

Step 5.2.2

Raise $2.71828182$ to the power of $0.5$ .

$f (0.5, 1.5) = 1.64872127$

Step 6

Use the recursive formula $y_{2} = y_{1} + h f (t_{1}, y_{1})$ to find $y_{2}$ .

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Step 6.1

Substitute.

$y_{2} = 1.5 + 0.5 \cdot 1.64872127$

Step 6.2

Simplify.

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Step 6.2.1

Multiply $0.5$ by $1.64872127$ .

$y_{2} = 1.5 + 0.82436063$

Step 6.2.2

Add $1.5$ and $0.82436063$ .

$y_{2} = 2.32436063$

Step 7

Use the recursive formula $t_{2} = t_{1} + h$ to find $t_{2}$ .

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Step 7.1

Substitute.

$t_{2} = 0.5 + 0.5$

Step 7.2

Add $0.5$ and $0.5$ .

$t_{2} = 1$

Step 8

Continue in the same manner until the desired values are approximated.

Step 9

List the approximations in a table.

$\begin{array}{c} t_{n} & y_{n} \\ 0 & 1 \\ 0.5 & 1.5 \\ 1 & 2.32436063 \\ 1.5 & 3.68350154 \\ 2 & 5.92434608 \\ 2.5 & 9.61887413 \\ 3 & 15.71012111 \\ 3.5 & 25.75288957 \\ 4 & 42.31061555 \\ 4.5 & 69.60969057 \\ 5 & 114.61825622 \end{array}$

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