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Posttest
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ODE: Runge Kutta method fourth order
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Beginner
Intermediate
Advanced
Runge Kutta method 4th order is used to solve
a: boundary value differential equation
Explanation
Explanation
b: initial value differential equation
Explanation
Explanation
c:
Explanation
Explanation
d:
Explanation
Explanation
The formulation used for Runge Kutta method 4th order is
a:
y
2
=
y
1
+
h
f
(
x
1
,
y
1
)
y_2 = y_1 + hf(x_1, y_1)
y
2
=
y
1
+
h
f
(
x
1
,
y
1
)
Explanation
Explanation
b:
y
2
=
h
f
(
x
1
,
y
1
)
y_2 = hf(x_1, y_1)
y
2
=
h
f
(
x
1
,
y
1
)
Explanation
Explanation
c:
s
1
=
h
f
(
x
1
,
y
1
)
s
2
=
h
f
(
x
1
+
h
,
y
1
+
s
1
)
y
2
=
y
1
+
(
s
1
+
s
2
)
2
\begin{aligned} s_1 &= hf(x_1, y_1) \\ s_2 &= hf(x_1+h, y_1 + s_1) \\ y_2 &= y_1 + \frac{(s_1 + s_2)}{2} \end{aligned}
s
1
s
2
y
2
=
h
f
(
x
1
,
y
1
)
=
h
f
(
x
1
+
h
,
y
1
+
s
1
)
=
y
1
+
2
(
s
1
+
s
2
)
Explanation
Explanation
d:
s
1
=
h
f
(
x
1
,
y
1
)
s
2
=
h
f
(
x
1
+
h
2
,
y
1
+
s
1
2
)
s
3
=
h
f
(
x
1
+
h
2
,
y
1
+
s
2
2
)
s
4
=
h
f
(
x
1
+
h
,
y
1
+
s
3
)
y
2
=
y
1
+
(
s
1
+
2
s
2
+
2
s
3
+
s
4
)
6
\begin{aligned} s_1 &= hf(x_1, y_1) \\ s_2 &= hf\left(x_1 + \frac{h}{2} , y_1 + \frac{s_1}{2}\right) \\ s_3 &= hf\left(x_1 + \frac{h}{2}, y_1 + \frac{s_2}{2}\right) \\ s_4 &= hf(x_1+h, y_1 + s_3) \\ y_2 &= y_1 + \frac{(s_1 + 2s_2 + 2s_3 + s_4)}{6} \end{aligned}
s
1
s
2
s
3
s
4
y
2
=
h
f
(
x
1
,
y
1
)
=
h
f
(
x
1
+
2
h
,
y
1
+
2
s
1
)
=
h
f
(
x
1
+
2
h
,
y
1
+
2
s
2
)
=
h
f
(
x
1
+
h
,
y
1
+
s
3
)
=
y
1
+
6
(
s
1
+
2
s
2
+
2
s
3
+
s
4
)
Explanation
Explanation
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