Calculus Examples

\frac{d y}{d t} = e^{t}

$\frac{d y}{d t} = e^{t}$ ,

y (0) = 0

$y (0) = 0$ ,

t = 1

$t = 1$ ,

h = 0.1

$h = 0.1$

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) = e^{t}

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

Step 2

Find

f (0, 0)

$f (0, 0)$ .

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Step 2.1

Substitute

0

$0$ for

t

$t$ and

0

$0$ for

y

$y$ .

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

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

Step 2.2

Simplify.

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Step 2.2.1

Replace

e

$e$ with an approximation.

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

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

Step 2.2.2

Raise

2.71828182

$2.71828182$ to the power of

0

$0$ .

f (0, 0) = 1

$f (0, 0) = 1$

f (0, 0) = 1

$f (0, 0) = 1$

f (0, 0) = 1

$f (0, 0) = 1$

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} = 0 + 0.1 \cdot 1

$y_{1} = 0 + 0.1 \cdot 1$

Step 3.2

Simplify.

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Step 3.2.1

Multiply

0.1

$0.1$ by

1

$1$ .

y_{1} = 0 + 0.1

$y_{1} = 0 + 0.1$

Step 3.2.2

Add

0

$0$ and

0.1

$0.1$ .

y_{1} = 0.1

$y_{1} = 0.1$

y_{1} = 0.1

$y_{1} = 0.1$

y_{1} = 0.1

$y_{1} = 0.1$

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 + 0.1

$t_{1} = 0 + 0.1$

Step 4.2

Add

0

$0$ and

0.1

$0.1$ .

t_{1} = 0.1

$t_{1} = 0.1$

t_{1} = 0.1

$t_{1} = 0.1$

Step 5

Find

f (0.1, 0.1)

$f (0.1, 0.1)$ .

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Step 5.1

Substitute

0.1

$0.1$ for

t

$t$ and

0.1

$0.1$ for

y

$y$ .

f (0.1, 0.1) = e^{0.1}

$f (0.1, 0.1) = e^{0.1}$

Step 5.2

Simplify.

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Step 5.2.1

Replace

e

$e$ with an approximation.

f (0.1, 0.1) = {2.71828182}^{0.1}

$f (0.1, 0.1) = {2.71828182}^{0.1}$

Step 5.2.2

Raise

2.71828182

$2.71828182$ to the power of

0.1

$0.1$ .

f (0.1, 0.1) = 1.10517091

$f (0.1, 0.1) = 1.10517091$

f (0.1, 0.1) = 1.10517091

$f (0.1, 0.1) = 1.10517091$

f (0.1, 0.1) = 1.10517091

$f (0.1, 0.1) = 1.10517091$

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} = 0.1 + 0.1 \cdot 1.10517091

$y_{2} = 0.1 + 0.1 \cdot 1.10517091$

Step 6.2

Simplify.

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Step 6.2.1

Multiply

0.1

$0.1$ by

1.10517091

$1.10517091$ .

y_{2} = 0.1 + 0.11051709

$y_{2} = 0.1 + 0.11051709$

Step 6.2.2

Add

0.1

$0.1$ and

0.11051709

$0.11051709$ .

y_{2} = 0.21051709

$y_{2} = 0.21051709$

y_{2} = 0.21051709

$y_{2} = 0.21051709$

y_{2} = 0.21051709

$y_{2} = 0.21051709$

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} = 0.1 + 0.1

$t_{2} = 0.1 + 0.1$

Step 7.2

Add

0.1

$0.1$ and

0.1

$0.1$ .

t_{2} = 0.2

$t_{2} = 0.2$

t_{2} = 0.2

$t_{2} = 0.2$

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 & 0 \\ 0.1 & 0.1 \\ 0.2 & 0.21051709 \\ 0.3 & 0.33265736 \\ 0.4 & 0.46764324 \\ 0.5 & 0.61682571 \\ 0.6 & 0.78169784 \\ 0.7 & 0.96390972 \\ 0.8 & 1.16528499 \\ 0.9 & 1.38783908 \\ 1 & 1.63379939 \end{array}

$\begin{array}{c} t_{n} & y_{n} \\ 0 & 0 \\ 0.1 & 0.1 \\ 0.2 & 0.21051709 \\ 0.3 & 0.33265736 \\ 0.4 & 0.46764324 \\ 0.5 & 0.61682571 \\ 0.6 & 0.78169784 \\ 0.7 & 0.96390972 \\ 0.8 & 1.16528499 \\ 0.9 & 1.38783908 \\ 1 & 1.63379939 \end{array}$

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