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Sum

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.

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#### Solution

(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 | x_{1} |
`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 | x_{2} |
`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

Concept: Modified Euler Method

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