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.

Enter a function: $$$y'=f(x,y)$$$ or $$$y'=f(t,y)=$$$

Enter the :

Enter the initial condition: $$$y$$$()$$$=$$$

Find $$$y$$$()

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Solution

Your input: find $$$y\left(1 \right)$$$ for $$$y'=t y$$$, when $$$y\left(0 \right)=3$$$, $$$h=\frac{1}{5}$$$ using the Euler's method.

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 $$$h=\frac{1}{5}$$$, $$$t_0=0$$$, $$$y_0=3$$$, $$$f(t,y)=t y$$$

Step 1.

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

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

$$$=3 + \frac{1}{5} \cdot \left(0 \right)=3$$$

Step 2.

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

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

$$$=3 + \frac{1}{5} \cdot \left(0.6 \right)=3.12$$$

Step 3.

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

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

$$$=3.12 + \frac{1}{5} \cdot \left(1.248 \right)=3.3696$$$

Step 4.

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

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

$$$=3.3696 + \frac{1}{5} \cdot \left(2.02176 \right)=3.773952$$$

Step 5.

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

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

$$$=3.773952 + \frac{1}{5} \cdot \left(3.0191616 \right)=4.37778432$$$

Answer: $$$y\left(1\right)=4.37778432$$$