I need the detailed steps for this solutions $($please write every step with explanation in organized way$)$

Apply the Runge$-$Kutta method to find the approximate value of $y$ for

$x=0\mathrm{.}2,$ in steps of $0\mathrm{.}1,$ if $d\frac{y}{d}x=x+y^2,y=1$ where $x=0.$

Solution:

Given $f(x,y)=x+y^2.$

Here we take $h=0\mathrm{.}1$ and carry out the calculations in two steps.

Step I. $x0=0,y0=1,h=0\mathrm{.}1$

:$.,k_1=hf(x_0,y_0)=0\mathrm{.}1f(0,1)=0\mathrm{.}1000$

$k_2=hf(x_0+\frac{1}{2}{h}_{,}{y}_0+\frac{1}{2}{k}_1)=0\mathrm{.}1f(0\mathrm{.}05,1\mathrm{.}1)=0\mathrm{.}1152$

$k_3=hf(x_0+\frac{1}{2}h,y_0+\frac{1}{2}{k}_2)=0\mathrm{.}1f(0\mathrm{.}05,1\mathrm{.}1152)=0\mathrm{.}1168$

$k_4=hf(x_0+h,y_0+k_3)=0\mathrm{.}1f(0\mathrm{.}1,1\mathrm{.}1168)=0\mathrm{.}1347$

:$.,k=\frac{1}{6}(k_1+2k_2+2k_3+k_4)$

$=\frac{1}{6}(0\mathrm{.}1000+0\mathrm{.}2304+0\mathrm{.}2336+0\mathrm{.}1347)=0\mathrm{.}1165$

giving $y(0\mathrm{.}1)=y_0+k=1\mathrm{.}1165$

Step $II.x_1=x_0+h=0\mathrm{.}1,y_1=1\mathrm{.}1165,h=0\mathrm{.}1$

:$.,k_1,=hf(x_1,y_1)=0\mathrm{.}1f(0\mathrm{.}1,1\mathrm{.}1165)=0\mathrm{.}1347$

$k_2,=hf(x_1+\frac{1}{2}h,y_1+\frac{1}{2}{k}_1)=0\mathrm{.}1f(0\mathrm{.}15,1\mathrm{.}1838)=0\mathrm{.}1551$

$,k_3=hf(x_1+\frac{1}{2}h,y_1+\frac{1}{2}{k}_2)=0\mathrm{.}1f(0\mathrm{.}15,1\mathrm{.}194)=0\mathrm{.}1576$

$,k_4,=hf(x_1+h,y_2+k_3)=0\mathrm{.}1f(0\mathrm{.}2,1\mathrm{.}1576)=0\mathrm{.}1823$

:$.,k,=\frac{1}{6}(k_1+2k_2+2k_3+k_4)=0\mathrm{.}1571$

Hence $y(0\mathrm{.}2)=y_1+k=1\mathrm{.}2736$