# Euler's Method Calculator

The calculator will find the approximate solution of the first-order differential equation using the Euler's method, with steps shown.

Or $y^{\prime } = f{\left(x,y \right)}$.
Or $x_{0}$.
$y_0=y(t_0)$ or $y_0=y(x_0)$.
Or $x_{1}$.

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Find $y{\left(1 \right)}$ for $y^{\prime } = t y$, when $y{\left(0 \right)} = 3$, $h = \frac{1}{5}$ using the Euler's method.

## Solution

The Euler's method states that $y_{n+1} = y_{n} + h\cdot f{\left(t_{n},y_{n} \right)}$, where $t_{n+1} = t_{n} + h$.

We have that $h = \frac{1}{5}$, $t_{0} = 0$, $y_{0} = 3$, and $f{\left(t,y \right)} = t y$.

### Step 1

$t_{1} = t_{0} + h = 0 + \frac{1}{5} = \frac{1}{5}$

$y_{1} = y{\left(t_{1} \right)} = y{\left(\frac{1}{5} \right)} = y_{0} + h\cdot f{\left(t_{0},y_{0} \right)} = 3 + h\cdot f{\left(0,3 \right)} = 3 + \frac{1}{5} \cdot 0 = 3$

### Step 2

$t_{2} = t_{1} + h = \frac{1}{5} + \frac{1}{5} = \frac{2}{5}$

$y_{2} = y{\left(t_{2} \right)} = y{\left(\frac{2}{5} \right)} = y_{1} + h\cdot f{\left(t_{1},y_{1} \right)} = 3 + h\cdot f{\left(\frac{1}{5},3 \right)} = 3 + \frac{1}{5} \cdot 0.6 = 3.12$

### Step 3

$t_{3} = t_{2} + h = \frac{2}{5} + \frac{1}{5} = \frac{3}{5}$

$y_{3} = y{\left(t_{3} \right)} = y{\left(\frac{3}{5} \right)} = y_{2} + h\cdot f{\left(t_{2},y_{2} \right)} = 3.12 + h\cdot f{\left(\frac{2}{5},3.12 \right)} = 3.12 + \frac{1}{5} \cdot 1.248 = 3.3696$

### Step 4

$t_{4} = t_{3} + h = \frac{3}{5} + \frac{1}{5} = \frac{4}{5}$

$y_{4} = y{\left(t_{4} \right)} = y{\left(\frac{4}{5} \right)} = y_{3} + h\cdot f{\left(t_{3},y_{3} \right)} = 3.3696 + h\cdot f{\left(\frac{3}{5},3.3696 \right)} = 3.3696 + \frac{1}{5} \cdot 2.02176 = 3.773952$

### Step 5

$t_{5} = t_{4} + h = \frac{4}{5} + \frac{1}{5} = 1$

$y_{5} = y{\left(t_{5} \right)} = y{\left(1 \right)} = y_{4} + h\cdot f{\left(t_{4},y_{4} \right)} = 3.773952 + h\cdot f{\left(\frac{4}{5},3.773952 \right)} = 3.773952 + \frac{1}{5} \cdot 3.0191616 = 4.37778432$

$y{\left(1 \right)}\approx 4.37778432$A