Respuesta :
Answer:
Therefore the value of y(1)= 0.9152.
Step-by-step explanation:
According to the Euler's method
y(x+h)≈ y(x) + hy'(x) ....(1)
Given that y(0) =3 and step size (h) = 0.2.
[tex]y'(x)= x^2y(x)-\frac12y^2(x)[/tex]
Putting the value of y'(x) in equation (1)
[tex]y(x+h)\approx y(x) +h(x^2y(x)-\frac12y^2(x))[/tex]
Substituting x =0 and h= 0.2
[tex]y(0+0.2)\approx y(0)+0.2[0\times y(0)-\frac12 (y(0))^2][/tex]
[tex]\Rightarrow y(0.2)\approx 3+0.2[-\frac12 \times3][/tex] [∵ y(0) =3 ]
[tex]\Rightarrow y(0.2)\approx 2.7[/tex]
Substituting x =0.2 and h= 0.2
[tex]y(0.2+0.2)\approx y(0.2)+0.2[(0.2)^2\times y(0.2)-\frac12 (y(0.2))^2][/tex]
[tex]\Rightarrow y(0.4)\approx 2.7+0.2[(0.2)^2\times 2.7- \frac12(2.7)^2][/tex]
[tex]\Rightarrow y(0.4)\approx 1.9926[/tex]
Substituting x =0.4 and h= 0.2
[tex]y(0.4+0.2)\approx y(0.4)+0.2[(0.4)^2\times y(0.4)-\frac12 (y(0.4))^2][/tex]
[tex]\Rightarrow y(0.6)\approx 1.9926+0.2[(0.4)^2\times 1.9926- \frac12(1.9926)^2][/tex]
[tex]\Rightarrow y(0.6)\approx 1.6593[/tex]
Substituting x =0.6 and h= 0.2
[tex]y(0.6+0.2)\approx y(0.6)+0.2[(0.6)^2\times y(0.6)-\frac12 (y(0.6))^2][/tex]
[tex]\Rightarrow y(0.8)\approx 1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2][/tex]
[tex]\Rightarrow y(0.6)\approx 0.8800[/tex]
Substituting x =0.8 and h= 0.2
[tex]y(0.8+0.2)\approx y(0.8)+0.2[(0.8)^2\times y(0.8)-\frac12 (y(0.8))^2][/tex]
[tex]\Rightarrow y(1.0)\approx 0.8800+0.2[(0.8)^2\times 0.8800- \frac12(0.8800)^2][/tex]
[tex]\Rightarrow y(1.0)\approx 0.9152[/tex]
Therefore the value of y(1)= 0.9152.
