Approximating Solutions Using Euler’s Method BC only

Euler's method chains tangent lines: from $(x_{n}, y_{n})$ the equation gives a slope, and $y_{n+1} = y_{n} + h \cdot f(x_{n}, y_{n})$ with $x_{n+1} = x_{n} + h$. For $\frac{dy}{dx} = x + y$ with $y(0) = 1$ and $h = 0.5$, the slope at $(0, 1)$ is $1$ so $y = 1.5$ at $x = 0.5$; the slope there is $2$ so $y = 2.5$ at $x = 1$. The step size is given directly or as $h = \frac{b - a}{n}$ for $n$ equal steps, it is negative when the target lies to the left, and the count matters as much as the size.

The slope must be recomputed at every new point using both updated coordinates; reusing the first slope turns the polygon into one long tangent line and gives $2$ rather than $2.5$ in the example above. Because each step follows a tangent, Euler's method underestimates where the solution is concave up and overestimates where it is concave down: here $\frac{d^{2}y}{dx^{2}} = 1 + x + y$ is positive, and the estimate $2.5$ falls short of the exact $2e - 2 \approx 3.437$. A smaller step is more accurate, and the error accumulates across steps.