Use Eulers method with step size 03 to estimate y15 where yx
Solution
Given differential equation : y\' = -2x + y^2,
f(x, y ) = -2x + y^2
and also given y(x) = y(0) = 1
x0 = 0 , y0 =1
f0 = f( 0, 1) = -2(0) + (1)^2
f0 = 1
And given step size h = 0.3
y1 = y0 + h f(x0 , y0 )
= 1 + 0.3 (1 )
y1 = 1.3
for next step , since h = 0.3 , the next point is x+ h = 0 + 0.3 = 0.3 ,we substitute what we know in euler\'s formula , and we have
y(x + h) = y(x) + h f(x, y )
y2 = y1 + h f(x1 , y1)
y2 = 1.3 + 0.3( -2 *0.3 + (1.3)^2)
= 1.3 + 0.3 (-0.6 + 1.69)
y2 = 1.627
now for third step ; x2 = 0.3 + 0.3 = 0.6 , y2 = 1.627
y3 = y2 + h f(x2 , y2 )
y3 = 1.627 + 0.3 (- 2 * 0.6 + (1.627)^2)
= 1.627 + 0.3 ( -1.2 + 2.6471)
= 2.061
y4 = y3 + h f( x3, y3) ; x3 = 0.6 + 0.3= 0.9
= 2.061 + 0.3 (- 2 * 0.9 + (2.061)^2)
= 2.061 + 0.3 ( -1.8 + 4.2477)
y4 = 2.795
y5 = y4 + h f(x4 , y4) , x4 = 0.9+0.3 = 1.2
= 2.795 + 0.3 ( -2 * 1.2 + (2.795)^2))
= 2.795 + 0.3 ( -2.4 + 7.812)
y5 = 4.4186
y6 = y5 + h f( x5 , y5) ; x5 = 1.2 +0.3 = 1.5
= 4.418 + 0.3 ( -2 * 1.5 + (4.418)^2)
= 4.418 + 0.3( -3 + 19.518)
= 9.3736
therefore y( 1.5 ) = 9.3736


