Question

Using Modified Eulers method ,find an approximate value of y At x = 0.2 in two step taking h=0.1 and using three iteration Given that `(dy)/(dx)=x+3y` , y = 1 when x = 0.

Answer 1

(I) `(dy)/(dx)=x+3y`  𝒙_𝟎=𝟎, 𝒚_𝟎=𝟏, 𝒉=𝟎.𝟏

`y_1^0=y_0+hf(x_0,y_0)=1+0.1(3)=1.3`

`y_1^(n+1)=y_0+h/2[f(x_0,y_0)+f(x_1,y_1^n)]`

Iteration	x1	`y_1^n`	`x_1y_1^n`	`y_1^(n+1)`
0	0.1	1.3	4	1.35
1	0.1	1.35	4.15	1.3575
2	0.1	1.3575	4.1725	1.3587

y(0.1)=1.3587

(II) 𝒙_𝟏=𝟎.𝟏,𝒚𝟏=𝟏.𝟑𝟓𝟖𝟕

`y_2^0=1.77631`

`y_2^(n+1)=y_1+h/2[f(x_2,y_2)+f(x_2,y_2^n)]`

Iteration	x2	`y_1^n`	`x_2y_2^n`	`y_2^(n+1)`
0	0.2	1.77631	5.52893	1.8439
1	0.2	1.8439	5.7317	1.8540
2	0.2	1.8540	5.762	1.8556

y(0.2)=1.8556