Approximate y(1,2), with h = 0.1 given that x y' = y -2x^2, y(1)=2, by RK-4 using 5-digits rounding.

Question
Upper Level Math
asked 2021-02-18
Approximate y(1,2), with h = 0.1 given that \(\displaystyle{x}{y}'={y}-{2}{x}^{{2}},{y}{\left({1}\right)}={2}\), by RK-4 using 5-digits rounding.

Answers (1)

2021-02-19
Step 1
To approximate y(1.2)by RK-4
Given h = 0.1 and \(\displaystyle{x}{y}'={y}-{2}{x}^{{2}},{y}{\left({1}\right)}={2}\)
\(\displaystyle{y}'=\frac{{y}}{{x}}-{2}{x}\)
Here \(\displaystyle{f{{\left({x},{y}\right)}}}=\frac{{y}}{{x}}-{2}{x}\)
Fourth order RK method is given by
\(\displaystyle{k}_{{1}}={h}{f{{\left({x}_{{0}},{y}_{{0}}\right)}}}\)
\(\displaystyle={\left({0.1}\right)}{f{{\left({1},{2}\right)}}}\)
\(\displaystyle={\left({0.1}\right)}\cdot{\left({0}\right)}={0}\)
\(\displaystyle{k}_{{2}}={h}{f{{\left({x}_{{0}}+\frac{{h}}{{2}},{y}_{{0}}+\frac{{k}_{{1}}}{{2}}\right)}}}\)
\(\displaystyle={\left({0.1}\right)}{f{{\left({1.05},{2}\right)}}}\)
\(\displaystyle={\left({0.1}\right)}\cdot{\left(-{0.19524}\right)}\)
\(\displaystyle=-{0.01952}\)
\(\displaystyle{k}_{{3}}={h}{f{{\left({x}_{{0}}+\frac{{h}}{{2}},{y}_{{0}}+\frac{{k}_{{2}}}{{2}}\right)}}}\)
\(\displaystyle={\left({0.1}\right)}{f{{\left({1.05},{1.99024}\right)}}}\)
\(\displaystyle={\left({0.1}\right)}\cdot{\left(-{0.20454}\right)}\)
=-0.02045
\(\displaystyle{k}_{{4}}={h}{f{{\left({x}_{{0}}+{h},{y}_{{0}}+{k}_{{3}}\right)}}}\)
\(\displaystyle={\left({0.1}\right)}{f{{\left({1.1},{1.97955}\right)}}}\)
\(\displaystyle={\left({0.1}\right)}\cdot{\left(-{0.40041}\right)}\)
=-0.04004
\(\displaystyle{y}_{{1}}={y}_{{0}}+\frac{{1}}{{6}}{\left({k}_{{1}}+{2}{k}_{{2}}+{2}{k}_{{3}}+{k}_{{4}}\right)}\)
\(\displaystyle{y}_{{1}}={2}+\frac{{1}}{{6}}{\left({0}+{2}{\left(-{0.01952}\right)}+{2}{\left(-{0.02045}\right)}+{\left(-{0.004004}\right)}\right)}\)
\(\displaystyle{y}_{{1}}={1.986}\)
Hence y(1.1)=1.98601
Step 2
Again taking \(\displaystyle{\left({x}_{{1}},{y}_{{1}}\right)}\in{p}{l}{a}{c}{e}{o}{f{{\left({x}_{{0}},{y}_{{0}}\right)}}}\) repeat the process
\(\displaystyle{k}_{{1}}={h}{f{{\left({x}_{{1}},{y}_{{1}}\right)}}}\)
\(\displaystyle={\left({0.1}\right)}{f{{\left({1.1},{1.98601}\right)}}}\)
\(\displaystyle={\left({0.1}\right)}\cdot{\left(-{0.394536}\right)}\)
=-0.0394536
\(\displaystyle{k}_{{2}}={h}{f{{\left({x}_{{1}}+\frac{{h}}{{2}},{y}_{{1}}+\frac{{k}_{{1}}}{{2}}\right)}}}\)
\(\displaystyle={\left({0.1}\right)}{f{{\left({1.15},{1.9662832}\right)}}}\)
\(\displaystyle{\left({0.1}\right)}\cdot{\left(-{0.590188}\right)}\)
=-0.0590188
\(\displaystyle{k}_{{3}}={h}{f{{\left({x}_{{1}}+\frac{{h}}{{2}},{y}_{{1}}+\frac{{k}_{{2}}}{{2}}\right)}}}\)
\(\displaystyle{\left({0.1}\right)}{f{{\left({1.15},{1.9565006}\right)}}}\)
\(\displaystyle={\left({0.1}\right)}\cdot{\left(-{0.598695130}\right)}\)
=-0.0598695
\(\displaystyle{k}_{{4}}={h}{f{{\left({x}_{{1}}+{h},{y}_{{1}}+{k}_{{3}}\right)}}}\)
\(\displaystyle={\left({0.1}\right)}{f{{\left({1.2},{1.9261405}\right)}}}\)
\(\displaystyle={\left({0.1}\right)}\cdot{\left(-{0.79488}\right)}\)
=-0.079488
\(\displaystyle{y}_{{2}}={y}_{{1}}+\frac{{1}}{{6}}{\left({k}_{{1}}+{2}{k}_{{2}}+{2}{k}_{{3}}+{k}_{{4}}\right)}\)
\(\displaystyle{y}_{{2}}={1.98}+\frac{{1}}{{6}}{\left[-{0.0394536}+{2}{\left(-{0.0590188}\right)}+{2}{\left(-{0.0598695}\right)}+{\left(-{0.079488}\right)}\right]}\)
\(\displaystyle{y}_{{2}}={1.92055}\)
Hence y(1.2)=1.92055
0

Relevant Questions

asked 2020-11-16
Advanced Math
A bipolar alkaline water electrolyzer stack module comprises 160 electrolytic cells that have an effective cell area of \(\displaystyle{2}{m}^{{2}}\). At nominal operation, the current density for a single cell of the electrolyzer stack is 0.40 \(\displaystyle\frac{{A}}{{c}}{m}^{{2}}\). The nominal operating temperature of the water electrolyzer stack is \(\displaystyle{70}^{\circ}\) C and pressure 1 bar. The voltage over a single electrolytic cell is 1.96 V at nominal load and 1.78 V at 50% of nominal load. The Faraday efficiency of the water electrolyzer stack is 95% at nominal current density, but at 50% of nominal load, the Faraday efficiency decreases to 80%.
(Give your answer to at least three significant digits.)
Calculate the nominal stack voltage:
Answer in V
Calculate the nominal stack current:
Answer in A
Calculate the nominal power on the water electrolyzer stack:
Answer in kW
asked 2021-03-01
Using the Mathematical Induction to prove that: \(\displaystyle{3}^{{{2}{n}}}-{1}\) is divisible by 4, whenever n is a positive integer.
asked 2021-01-02
Find a transformation from u,v space to x,y,z space that takes the triangle \(\displaystyle{U}={\left[\begin{array}{cc} {0}&{0}\\{1}&{0}\\{0}&{1}\end{array}\right]}\) to the triangle
\(\displaystyle{T}={\left[{\left({1},{0}\right)},-{2}\right)},{\left(-{1},{2},{0}\right)},{\left({1},{1},{2}\right)}{]}\)
asked 2021-01-02
The rate of change of the volume of a snowball that is melting is proportional to the surface area of the snowball. Suppose the snowball is perfectly spherical. Then the volume (in centimeters cubed) of a ball of radius r centimeters is \(\displaystyle\frac{{4}}{{3}}\pi{r}^{{3}}\). The surface area is \(\displaystyle{4}\pi{r}^{{2}}\).Set up the differential equation for how r is changing. Then, suppose that at time t = 0 minutes, the radius is 10 centimeters. After 5 minutes, the radius is 8 centimeters. At what time t will the snowball be completely melted.
asked 2021-03-08
\(\displaystyle{\left({x}^{{4}}{y}^{{5}}\right)}^{{\frac{{1}}{{4}}}}{\left({x}^{{8}}{y}^{{5}}\right)}^{{\frac{{1}}{{5}}}}={x}^{{\frac{{j}}{{5}}}}{y}^{{\frac{{k}}{{4}}}}\)
In the equation above, j and k are constants. If the equation is true for all positive real values of x and y, what is the value of j - k?
A)3
B)4
C)5
D)6
asked 2021-01-31
The function W defined by \(\displaystyle{W}{\left(\alpha\right)}={1.9}\alpha+{2.3}\) models the weight of a kitten, in ounces, that is a weeks old for \(\displaystyle{0}\le\alpha\le{6}\). According to this model, what is the weight, in ounces, of a kitten that is 3 weeks old?
asked 2021-02-05
Using "Proof by Contraposition",sho that: if n is any odd integer and m is any even integer, then, \(\displaystyle{3}{m}^{{3}}+{2}{m}^{{2}}\) is odd.
asked 2020-11-01
Given the following function: \(\displaystyle{f{{\left({x}\right)}}}={1.01}{e}^{{{4}{x}}}-{4.62}{e}^{{{3}{x}}}-{3.11}{e}^{{{2}{x}}}+{12.2}{e}^{{{x}}}-{1.99}\) a)Use three-digit rounding frithmetic, the assumption that \(\displaystyle{e}^{{{1.53}}}={4.62}\), and the fact that \(\displaystyle{e}^{{{n}{x}}}={\left({e}^{{{x}}}\right)}^{{{n}}}\) to evaluate \(\displaystyle{f{{\left({1.53}\right)}}}\) b)Redo the same calculation by first rewriting the equation using the polynomial factoring technique c)Calculate the percentage relative errors in both part a) and b) to the true result \(\displaystyle{f{{\left({1.53}\right)}}}=-{7.60787}\)
asked 2020-10-19
Given the following function:
\(\displaystyle{f{{\left({x}\right)}}}={1.01}{e}^{{{4}{x}}}-{4.62}{e}^{{{3}{x}}}-{3.11}{e}^{{{2}{x}}}+{12.2}{e}^{{{x}}}\)
a) Use three-digit rounding frithmetic, the assumption that \(\displaystyle{e}^{{{1.53}}}={4.62}\), and the fact that \(\displaystyle{e}^{{{n}{x}}}={\left({e}^{{{x}}}\right)}^{{{n}}}\) to evaluate \(\displaystyle{f{{\left({1.53}\right)}}}\)
b) Redo the same calculation by first rewriting the equation using the polynomial factoring technique
c) Calculate the percentage relative errors in both part a) and b) to the true result \(\displaystyle{f{{\left({1.53}\right)}}}=-{7.60787}\)
asked 2021-01-28
Given \(f(x)\ \cos(2x)\) and the interval
\([0,\ 3\ \frac{\pi}{4}]\)
approximate the area bounded by the graph of f(x) and the axis on the interval using a left, right, and mid point Riemann sum with \(n = 3\)
...