#import "@preview/physica:0.9.6": *


== a)

$
  cases(
    5y + 2y' + y'' + 3 & = 0,
    y(0) & = 1,
    y'(0) & = 4
  )
$

we introduce the substitutions $y_1 = y$ and $y_2 = y'_1 = y$ such that we get
$
  5y_1 + 2y_2 + y_2 ' + 3 = 0 \
  => y_2 ' = y'' = -3 - 2y_2 - 5y_1
$
for the first equation. this yields an autonomous first-order system
$
  cases(
    y_1 ' = y_2,
    y_2 ' = -3 - 2y_2 - 5y_1
  )
$
since there is no explicit $t$ on the right sides of the equations, i.e. the
state of the system only relies on its current state.

since we only use a step size of $h = 1$, we simply do
$
  y_(n + 1) = y_n + y'_n,
$
so starting at $t = 0$ we get, using our initial values
$
  cases(
    y_1(0) = 1 quad #text(gray)[initial],
    y_2(0) = 4 quad #text(gray)[initial],
    y'_1(0) = y_2 (0) = 4,
    y'_2(0) = -3 - 2y_2(0) - 5y_1(0) = -16
  )
$
which we can plug in to get our next values for $y_1$ and $y_2$
$
  cases(
    y_1(1) = y_1(0) + y'_1(0) = 5,
    y_2(1) = -12,
  )
$
which concludes the first step in euler's method.

== b)

for $k = 0, 1, 2, 3$
$
  cases(
    y + 3y'' + 4y^((3)) + sin(t) & = 3,
    y^((k))(0) & = 4 - k
  )
$
yields five equations.

in this case, we can make yet more substitutions
$
  cases(
    y_0 & = y,
    y'_k & = y_(k+1) = y^((k+1))
  )
$
such that
$
     & cases(
         y_0 + 3 y_2 + 4 y_3 + sin(t) = 3,
         y_k (0) = 4 - k
       ) \
  => & cases(
         y'_0 = y_1,
         y'_1 = y_2,
         y'_2 = y_3,
         y'_3 = -1/4 (y_1 + 3y_3 + cos(t))
       )
$

so starting at $t = 0$ with step size $h = 1$ we obtain
$
  y_0(0) = 4, quad
  y_1(0) = 3, quad
  y_2(0) = 2, quad
  y_3(0) = 1
$
and
$
  y'_0(0) = 3, quad
  y'_1(0) = 2, quad
  y'_2(0) = 1, quad
  y'_3(0) = -7 slash 4
$
such that after one step we have
$
  cases(
    y_0(1) = y_0(0) + y'_0(1) = 7,
    y_1(1) = 5,
    y_2(1) = 3,
    y_3(1) = -3 slash 4,
  )
$

== c)

we have the system of ODEs
$
  cases(
    2u + cos(u) sin(u') + 3(u'')^3 + ln(t + 1) & = 42,
    v^2 - 2v v' + (v'')^2 + v^((3)) & = t,
    u^((k))(0) & = k^2\, quad #$k = 0, 1$,
    v^((k))(1) & = 1\, quad #$k = 0, 1, 2$
  )
$

this time we have two functions of $t$, $u$ and $v$. we can substitute as
usual, then each one independently, since the equations don't overlap.

let
$
  u^((k)) = u_k quad "and" quad v^((k)) = v_k
$
such that we obtain the two disjoint systems
$
      & cases(
          2u_0 + cos(u_0) sin(u_1) + 3(u_2)^3 + ln(t + 1) & = 42,
          u_k (0) & = k^2\, quad #$k = 0, 1$,
        ) \
  and & cases(
          v_0^2 - 2v_0 v_1 + v_2^2 + v_3 & = t,
          v_k (1) & = 1\, quad #$k = 0, 1, 2$
        )
$
which yields
$
      & cases(
          u'_0 = u_1,
          u'_1 = u_2,
          u'_2 = dv(, t) root(3, 1/3 (42 - 2u_0 - cos(u_0) sin(u_1) - ln(t + 1)))
        ) \
  and & cases(
          v'_0 = v_1,
          v'_1 = v_2,
          v'_2 = v_3,
          v'_3 = dv(, t) (t - v_0^2 + 2 v_0 v_1 - v_2^2)
        )
$
combined with the initial conditions
$
  cases(
    u_0 (0) = 0,
    u_1 (0) = 1
  ) quad
  and quad cases(
    v_0 (1) = 1,
    v_1 (1) = 1,
    v_2 (1) = 1
  )
$
yields for step size $h = 1$
$
  cases(
    u_0(1) & = u_0(0) + u_1(0) = 1,
    u_1(1) & = u_1(0) + u'_1(0) \
    & = 1 + root(3, 1/3 (42 - 2 u_0(0) - cos(u_0(0)) sin(u_1(0)) - ln(0 + 1))) \
    & = 1 + root(3, (42 - sin(1))/3) approx #{
      calc.round(1 + calc.root((42 - calc.sin(1)) / 3, 3), digits: 4)
    }
  )
$
and
$
  cases(
    v_0(2) & = v_0(1) + v_1(1) = 2,
    v_1(2) & = v_1(1) + v_2(1) = 2,
    v_2(2) & = v_2(1) + v'_2(1) \
           & = 1 + (1 - (v_0(1))^2 + 2 v_0(1) v_1(1) - (v_2(1))^2) \
           & = 2
  )
$

thus have I done a step of euler's method for the system of ODEs.