Calculus Examples Differential Equations

Approximate Using Euler's Method

$\frac{d y}{d t} = e^{t}$ , $y (0) = 0$ , $t = 1$ , $h = 0.1$

Step 1

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

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

Step 2

Find $f (0, 0)$ .

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Step 2.1

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

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

Step 2.2

Simplify.

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Step 2.2.1

Replace $e$ with an approximation.

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

Step 2.2.2

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

$f (0, 0) = 1$

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.

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

Step 3.2

Simplify.

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Step 3.2.1

Multiply $0.1$ by $1$ .

$y_{1} = 0 + 0.1$

Step 3.2.2

Add $0$ and $0.1$ .

$y_{1} = 0.1$

Step 4

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

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Step 4.1

Substitute.

$t_{1} = 0 + 0.1$

Step 4.2

Add $0$ and $0.1$ .

$t_{1} = 0.1$

Step 5

Find $f (0.1, 0.1)$ .

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Step 5.1

Substitute $0.1$ for $t$ and $0.1$ for $y$ .

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

Step 5.2

Simplify.

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Step 5.2.1

Replace $e$ with an approximation.

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

Step 5.2.2

Raise $2.71828182$ to the power of $0.1$ .

$f (0.1, 0.1) = 1.10517091$

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.

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

Step 6.2

Simplify.

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Step 6.2.1

Multiply $0.1$ by $1.10517091$ .

$y_{2} = 0.1 + 0.11051709$

Step 6.2.2

Add $0.1$ and $0.11051709$ .

$y_{2} = 0.21051709$

Step 7

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

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Step 7.1

Substitute.

$t_{2} = 0.1 + 0.1$

Step 7.2

Add $0.1$ and $0.1$ .

$t_{2} = 0.2$

Step 8

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

Step 9

List the approximations in a table.

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

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