“solve” → find \( y = \text{function of } x \) → bunch of points:
If equation is separable, linear, homogeneous, or exact we can solve exactly.
But if not or if we don't want to solve using the techniques (e.g., integration too messy), we can use numerical methods to find \( y \) given an \( x \).
Want \( y \), we know its slope at \( (x, y) \).
\( y \rightarrow \) solution we don't have
Given: \( y' = f(x, y) \)
\( (x_0, y_0) \) initial condition
find: \( y(x) \)
specified
decide step size: interval between \( x \)'s \( \rightarrow \) call it \( h \)
build tangent line at \( (x_0, y_0) \)
\( y - y_0 = f(x_0, y_0)(x - x_0) \quad \longrightarrow \quad y = y_0 + f(x_0, y_0)h \)
this gives \( y_1 \):
\( y_1 = y_0 + f(x_0, y_0)h \)
if \( x_1 \) is not the target \( x \), repeat
\( x_{n+1} = x_n + h \)
\( y_{n+1} = y_n + f(x_n, y_n)h \)
iterate until
\( x_{n+1} = \text{target } x \)
\( y' = 2y - 3x \)
\( y(0) = 1 \)
Note: \( f(x, y) = 2y - 3x \)
estimate \( y(0.5) \) using step size \( h = 0.25 \)
\( x_0 = 0 \)
\( y_0 = 1 \)
\( x_1 = x_0 + h = 0 + 0.25 = 0.25 \)
\( y_1 = y_0 + f(x_0, y_0)h \)
\( = 1 + [2(1) - 3(0)](0.25) = 1.5 \)
use "old" \( x, y \)
\( x_1 \) is not yet the target \( x \) (0.5), so repeat
\( x_2 = x_1 + h = 0.25 + 0.25 = 0.5 \) (target, so stop after this iteration)
\( y_2 = y_1 + f(x_1, y_1)h \)
\( = 1.5 + [2(1.5) - 3(0.25)](0.25) = \)
estimate of \( y(0.5) \)
Let's solve \( y' = 2y - 3x \), \( y(0) = 1 \) exactly and compare.
linear homogeneous
true \( y(0.5) = 2.1796 \) (estimate was 2.0625)
To improve estimate, use more steps or smaller step size.
If we had used \( h = 0.01 \) (50 steps) \( y(0.5) \approx 2.1729 \)
Generally, smaller \( h \rightarrow \) better estimate (more steps).
Usually we don't know the true \( y \) (if we had it why estimate?).
If further refinement of step size does not result in significant changes.
\( y' = 2y - 3x \), \( y(0) = 1 \), estimate \( y(1) \)