Calculus Examples

\frac{d y}{d t} = \sin (t)

$\frac{d y}{d t} = \sin (t)$ ,

y (0) = 1

$y (0) = 1$ ,

t = 0.5

$t = 0.5$ ,

h = 0.05

$h = 0.05$

Step 1

Define

f (t, y)

$f (t, y)$ such that

\frac{d y}{d t} = f (t, y)

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

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

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

Step 2

Find

f (0, 1)

$f (0, 1)$ .

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Step 2.1

Substitute

0

$0$ for

t

$t$ and

1

$1$ for

y

$y$ .

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

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

Step 2.2

Evaluate

\sin (0)

$\sin (0)$ .

f (0, 1) = 0

$f (0, 1) = 0$

f (0, 1) = 0

$f (0, 1) = 0$

Step 3

Use the recursive formula

y_{1} = y_{0} + h f (t_{0}, y_{0})

$y_{1} = y_{0} + h f (t_{0}, y_{0})$ to find

y_{1}

$y_{1}$ .

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Step 3.1

Substitute.

y_{1} = 1 + 0.05 \cdot 0

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

Step 3.2

Simplify.

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Step 3.2.1

Multiply

0.05

$0.05$ by

0

$0$ .

y_{1} = 1 + 0

$y_{1} = 1 + 0$

Step 3.2.2

Add

1

$1$ and

0

$0$ .

y_{1} = 1

$y_{1} = 1$

y_{1} = 1

$y_{1} = 1$

y_{1} = 1

$y_{1} = 1$

Step 4

Use the recursive formula

t_{1} = t_{0} + h

$t_{1} = t_{0} + h$ to find

t_{1}

$t_{1}$ .

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Step 4.1

Substitute.

t_{1} = 0 + 0.05

$t_{1} = 0 + 0.05$

Step 4.2

Add

0

$0$ and

0.05

$0.05$ .

t_{1} = 0.05

$t_{1} = 0.05$

t_{1} = 0.05

$t_{1} = 0.05$

Step 5

Find

f (0.05, 1)

$f (0.05, 1)$ .

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Step 5.1

Substitute

0.05

$0.05$ for

t

$t$ and

1

$1$ for

y

$y$ .

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

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

Step 5.2

Evaluate

\sin (0.05)

$\sin (0.05)$ .

f (0.05, 1) = 0.04997916

$f (0.05, 1) = 0.04997916$

f (0.05, 1) = 0.04997916

$f (0.05, 1) = 0.04997916$

Step 6

Use the recursive formula

y_{2} = y_{1} + h f (t_{1}, y_{1})

$y_{2} = y_{1} + h f (t_{1}, y_{1})$ to find

y_{2}

$y_{2}$ .

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Step 6.1

Substitute.

y_{2} = 1 + 0.05 \cdot 0.04997916

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

Step 6.2

Simplify.

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Step 6.2.1

Multiply

0.05

$0.05$ by

0.04997916

$0.04997916$ .

y_{2} = 1 + 0.00249895

$y_{2} = 1 + 0.00249895$

Step 6.2.2

Add

1

$1$ and

0.00249895

$0.00249895$ .

y_{2} = 1.00249895

$y_{2} = 1.00249895$

y_{2} = 1.00249895

$y_{2} = 1.00249895$

y_{2} = 1.00249895

$y_{2} = 1.00249895$

Step 7

Use the recursive formula

t_{2} = t_{1} + h

$t_{2} = t_{1} + h$ to find

t_{2}

$t_{2}$ .

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Step 7.1

Substitute.

t_{2} = 0.05 + 0.05

$t_{2} = 0.05 + 0.05$

Step 7.2

Add

0.05

$0.05$ and

0.05

$0.05$ .

t_{2} = 0.1

$t_{2} = 0.1$

t_{2} = 0.1

$t_{2} = 0.1$

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}

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