Question

If 𝒚 satisfies the equation `(dy)/(dx)=x^2y-1` with `x_0=0, y_0=1` using Taylor’s Series Method find 𝒚 𝒂𝒕 𝒙= 𝟎.𝟏 (take h=0.1).

Answer 1

`(dy)/(dx)=x^2y-1` `x_0=0, y_0=1` 𝒉=𝟎.𝟏
To find : 𝒚(𝟎.𝟏)

`y'=x^2y-1  ,  y_0'=-1`

`y''=x^2y'+2xy  ,  y_0''=0`

`y'''=x^2y''+2y'x+2y+2xy  ,  y_0'''=0`

Taylor’s series is :

`y=y_0+h.y_0'+h^2/(2!)y_0''+h^3/(3!)y_0'''+...`

∴ 𝒚(𝟎.𝟏)=𝟏+𝟎.𝟏(−𝟏)+𝟎+`(0.1)^3/(3!)(2)`

∴ 𝒚(𝟎.𝟏)=𝟎.𝟗𝟎𝟎𝟑