# 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
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.

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

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