Question

Question: Use Euler's method with h = 0.1 to approximate x(0.2) and y(0.2), where x(t), y(t) is the solution x' = x + y of the initial value problem y' = x - y and x(0) = 1, y(0) = 2.

Short answer

So, the approximate values are $x\left(0.2\right)=1.62,y\left(0.2\right)=1.84$.

Step-by-step solution

Definition of Euler's method

The Euler method is a first-order method, which means that the local error (error per step) is proportional to the square of the step size and the global error (error at a given time) is proportional to the step size.

Use Euler’s method

Given system is$x^{\text{'}}=x+y$

$y^{\text{'}}=x-yx\left(0\right)=1,\left(0\right)=2$

So $f_1(x,y)=x+y,f_2(x,y)=x-y$Let $h=0.1$

Applying Euler's method successively we get

$\begin{array}{rcl}{x}_1& =& x_0+h{f}_1\left(x_0,y_0\right)\\ & =& x_0+h\left(x_0+y_0\right)\\ & =& 1+0.1\left(1+2\right)\\ x_1& =& 1.3\\ & & \\ y_1& =& y_0+h{f}_2\left(x_0,y_0\right)\\ & =& y_0+h\left(x_0-y_0\right)\\ & =& 2+0.1(1-2)\\ y_1& =& 1.9\end{array}$

Evaluation

Now for second time

$\begin{array}{rcl}{x}_2& =& x_1+h{f}_1\left(x_1,y_1\right)\\ & =& 1.3+0.1[1.3+1.9]\\ & =& 1.69\\ & & \\ {\text{y}}_2& =& {\text{y}}_1+h{\text{f}}_2\left({\text{x}}_1,{\text{y}}_1\right)\\ & =& 1.9+0.1[1.3-1.9]\\ {\text{y}}_2& =& 1.84\end{array}$

Thus, the required values are $x\left(0.2\right)=1.62,y\left(0.2\right)=1.84$.