 # 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. Emeli Hagan 2021-02-18 Answered
Approximate y(1,2), with h = 0.1 given that $x{y}^{\prime }=y-2{x}^{2},y\left(1\right)=2$, by RK-4 using 5-digits rounding.
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Step 1
To approximate y(1.2)by RK-4
Given $h=0.1$ and $x{y}^{\prime }=y-2{x}^{2},y\left(1\right)=2$
${y}^{\prime }=\frac{y}{x}-2x$
Here $f\left(x,y\right)=\frac{y}{x}-2x$
Fourth order RK method is given by
${k}_{1}=hf\left({x}_{0},{y}_{0}\right)$
$=\left(0.1\right)f\left(1,2\right)$
$=\left(0.1\right)\cdot \left(0\right)=0$
${k}_{2}=hf\left({x}_{0}+\frac{h}{2},{y}_{0}+\frac{{k}_{1}}{2}\right)$
$=\left(0.1\right)f\left(1.05,2\right)$
$=\left(0.1\right)\cdot \left(-0.19524\right)$
$=-0.01952$
${k}_{3}=hf\left({x}_{0}+\frac{h}{2},{y}_{0}+\frac{{k}_{2}}{2}\right)$
$=\left(0.1\right)f\left(1.05,1.99024\right)$
$=\left(0.1\right)\cdot \left(-0.20454\right)$
$=-0.02045$
${k}_{4}=hf\left({x}_{0}+h,{y}_{0}+{k}_{3}\right)$
$=\left(0.1\right)f\left(1.1,1.97955\right)$
$=\left(0.1\right)\cdot \left(-0.40041\right)$
$=-0.04004$
${y}_{1}={y}_{0}+\frac{1}{6}\left({k}_{1}+2{k}_{2}+2{k}_{3}+{k}_{4}\right)$
${y}_{1}=2+\frac{1}{6}\left(0+2\left(-0.01952\right)+2\left(-0.02045\right)+\left(-0.004004\right)\right)$
${y}_{1}=1.986$
Hence $y\left(1.1\right)=1.98601$
Step 2
Again taking $\left({x}_{1},{y}_{1}\right)\in placeof\left({x}_{0},{y}_{0}\right)$ repeat the process
${k}_{1}=hf\left({x}_{1},{y}_{1}\right)$
$=\left(0.1\right)f\left(1.1,1.98601\right)$

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