Calculus Examples

$\frac{d y}{d t} = \sin (t)$ , $y (0) = 1$ , $t = 0.5$ , $h = 0.05$

Step 1

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

$f (t, y) = \sin (t)$

Step 2

Find $f (0, 1)$ .

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Step 2.1

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

$f (0, 1) = \sin (0)$

Step 2.2

Evaluate $\sin (0)$ .

$f (0, 1) = 0$

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.

$y_{1} = 1 + 0.05 \cdot 0$

Step 3.2

Simplify.

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Step 3.2.1

Multiply $0.05$ by $0$ .

$y_{1} = 1 + 0$

Step 3.2.2

Add $1$ and $0$ .

$y_{1} = 1$

Step 4

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

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Step 4.1

Substitute.

$t_{1} = 0 + 0.05$

Step 4.2

Add $0$ and $0.05$ .

$t_{1} = 0.05$

Step 5

Find $f (0.05, 1)$ .

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Step 5.1

Substitute $0.05$ for $t$ and $1$ for $y$ .

$f (0.05, 1) = \sin (0.05)$

Step 5.2

Evaluate $\sin (0.05)$ .

$f (0.05, 1) = 0.04997916$

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.

$y_{2} = 1 + 0.05 \cdot 0.04997916$

Step 6.2

Simplify.

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Step 6.2.1

Multiply $0.05$ by $0.04997916$ .

$y_{2} = 1 + 0.00249895$

Step 6.2.2

Add $1$ and $0.00249895$ .

$y_{2} = 1.00249895$

Step 7

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

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Step 7.1

Substitute.

$t_{2} = 0.05 + 0.05$

Step 7.2

Add $0.05$ and $0.05$ .

$t_{2} = 0.1$

Step 8

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

Step 9

List the approximations in a table.

$\begin{array}{c} t_{n} & y_{n} \\ 0 & 1 \\ 0.05 & 1 \\ 0.1 & 1.00249895 \\ 0.15 & 1.00749062 \\ 0.2 & 1.01496253 \\ 0.25 & 1.024896 \\ 0.3 & 1.0372662 \\ 0.35 & 1.05204221 \\ 0.4 & 1.0691871 \\ 0.45 & 1.08865801 \\ 0.5 & 1.11040629 \end{array}$

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