Consider that math modeling following initial valu problem dy/dt=3-2t-0.5y, y(0)=1

Given Initial Value Problem is:
${\displaystyle \frac{dy}{dt}=3-2t-0,5y,y\left(0\right)=1,t_0=0}$
and step size $h=0,05$
Here ${\displaystyle f(t,y)=3-2t-0,5y}$
${\displaystyle f(t_{n-1},y_{n-1})=3-2t_{n-1}-0,5y_{n-1}}$
${\displaystyle =3-2\{t_0+(n-1)h\}-0,5y_{n-1}}$
Now we have ${\displaystyle t_0=0\text{and}{y}_0=1}$
for $n=1$ ${\displaystyle \therefore f(t_0,y_0)=3-2\{0+(1-1)h\}-0,5}$
${\displaystyle =3-2\times 0-0,5\times 1}$
$=3-0,5$
$=2,5$
${\displaystyle \therefore y_1=y_0+hf(t_0,y_0)}$
${\displaystyle =1+0,05\times 2,5}$
$=1+0,125$
$=1,125$
Now, ${\displaystyle t_1=t_0+(n-1)h=0+(2-1)\times 0,05=0,05,f_{0|><}n=2f_{0|><}}$
$n=2,$
${\displaystyle f(t_1,y_1)=3-2\{t_0+(n-1)h\}-0,5y_1}$
${\displaystyle =3-2\{0+(2-1)0,05\}-0,5\times 1,125}$
${\displaystyle =3-2\times 0,05-0,5625}$
${\displaystyle =3-0,1-0,05625}$
$=2,3375$
${\displaystyle \therefore y_2=y_1=hf(t_1,y_1)}$
${\displaystyle =1,125+0,05\times 2,3375}$

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