Modified Euler's Method Calculator

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

$$$y^{\prime }\left(t\right) = f{\left(t,y \right)}$$$:

Or $$$y^{\prime }\left(x\right) = f{\left(x,y \right)}$$$.

Divide using:

$$$h$$$:

$$$t_{0}$$$:

Or $$$x_{0}$$$.

$$$y_{0}$$$:

$$$y_0=y(t_0)$$$ or $$$y_0=y(x_0)$$$.

$$$t_{1}$$$:

Or $$$x_{1}$$$.

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Your Input

Find $$$y{\left(1 \right)}$$$ for $$$y^{\prime }\left(t\right) = 2 t - y$$$, when $$$y{\left(0 \right)} = 1$$$, $$$h = \frac{1}{5}$$$ using the modified Euler's method.

Solution

The modified Euler's method states that $$$y_{n+1} = y_{n} + h f{\left(t_{n} + \frac{h}{2},y_{n} + \frac{h}{2} f{\left(t_{n},y_{n} \right)} \right)}$$$, where $$$t_{n+1} = t_{n} + h$$$.

We have that $$$h = \frac{1}{5}$$$, $$$t_{0} = 0$$$, $$$y_{0} = 1$$$, and $$$f{\left(t,y \right)} = 2 t - y$$$.

Step 1

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

$$$f{\left(t_{0},y_{0} \right)} = f{\left(0,1 \right)} = -1$$$

$$$y_{1} = y{\left(t_{1} \right)} = y{\left(\frac{1}{5} \right)} = y_{0} + h f{\left(t_{0} + \frac{h}{2},y_{0} + \frac{h}{2} f{\left(t_{0},y_{0} \right)} \right)} = 1 + \frac{f{\left(0 + \frac{\frac{1}{5}}{2},1 + \frac{\frac{1}{5}}{2} \left(-1\right) \right)}}{5} = 0.86$$$

Step 2

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

$$$f{\left(t_{1},y_{1} \right)} = f{\left(\frac{1}{5},0.86 \right)} = -0.46$$$

$$$y_{2} = y{\left(t_{2} \right)} = y{\left(\frac{2}{5} \right)} = y_{1} + h f{\left(t_{1} + \frac{h}{2},y_{1} + \frac{h}{2} f{\left(t_{1},y_{1} \right)} \right)} = 0.86 + \frac{f{\left(\frac{1}{5} + \frac{\frac{1}{5}}{2},0.86 + \frac{\frac{1}{5}}{2} \left(-0.46\right) \right)}}{5} = 0.8172$$$

Step 3

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

$$$f{\left(t_{2},y_{2} \right)} = f{\left(\frac{2}{5},0.8172 \right)} = -0.0172$$$

$$$y_{3} = y{\left(t_{3} \right)} = y{\left(\frac{3}{5} \right)} = y_{2} + h f{\left(t_{2} + \frac{h}{2},y_{2} + \frac{h}{2} f{\left(t_{2},y_{2} \right)} \right)} = 0.8172 + \frac{f{\left(\frac{2}{5} + \frac{\frac{1}{5}}{2},0.8172 + \frac{\frac{1}{5}}{2} \left(-0.0172\right) \right)}}{5} = 0.854104$$$

Step 4

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

$$$f{\left(t_{3},y_{3} \right)} = f{\left(\frac{3}{5},0.854104 \right)} = 0.345896$$$

$$$y_{4} = y{\left(t_{4} \right)} = y{\left(\frac{4}{5} \right)} = y_{3} + h f{\left(t_{3} + \frac{h}{2},y_{3} + \frac{h}{2} f{\left(t_{3},y_{3} \right)} \right)} = 0.854104 + \frac{f{\left(\frac{3}{5} + \frac{\frac{1}{5}}{2},0.854104 + \frac{\frac{1}{5}}{2} \cdot 0.345896 \right)}}{5} = 0.95636528$$$

Step 5

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

$$$f{\left(t_{4},y_{4} \right)} = f{\left(\frac{4}{5},0.95636528 \right)} = 0.64363472$$$

$$$y_{5} = y{\left(t_{5} \right)} = y{\left(1 \right)} = y_{4} + h f{\left(t_{4} + \frac{h}{2},y_{4} + \frac{h}{2} f{\left(t_{4},y_{4} \right)} \right)} = 0.95636528 + \frac{f{\left(\frac{4}{5} + \frac{\frac{1}{5}}{2},0.95636528 + \frac{\frac{1}{5}}{2} \cdot 0.64363472 \right)}}{5} = 1.1122195296$$$

Answer

$$$y{\left(1 \right)}\approx 1.1122195296$$$A

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