# Approximate the solution around `x_0=0` with a fourth order taylor polynomial for the following initial value problem. `y''=3y'+x^(7/3)y, y(0)=10, y'(0)=5.` A fourth order taylor polynomial `p_4(x)` about `x_0=0` has the form

`p_4(x)=y(x_0)+y'(x_0)x+(y''(x_0))/(2!)x^2+(y'''(x_0)) /(3!)x^3+(y^((4))(x_0))/(4!) x^4`

We need the values `y(0)` , `y'(0)` , `y''(0)` , etc. The first two are provided for us and we can use the differential equation itself to find the others.

Plug `x=x_0=0` into the equation and solve for `y''(0)` .

`y''(0)=3y'(0)+0^(7/3)y(0)=3*5+0=15`

Differentiate the equation on both sides now to find `y'''(0)` and `y^((4))(0)` .

`d/(dx)[y''(x)]=d/(dx)[3y'(x)+x^(7/3)y(x)]`

`y'''=3y''+(7/3)x^(4/3)y+x^(7/3)y''`

Now once more.

`y^((4))=3y'''+(28/9)x^(1/3)y+(14/13)x^(4/3)y'+x^(7/3)y''`

Plug in `x=0` to evaluate `y'''(0)` and `y^((4))(0)` .

`y'''(0)=3y''(0)+(7/3)(0)^(4/3)y(0)+(0)^(7/3)y''(0)`

`y'''(0)=3y''(0)=3*15=45`

`y^((4))(0)=3y'''(0)+(28/9)(0)^(1/3)y+(14/13)(0)^(4/3)y'(0)+(0)^(7/3)(0)''`

`y^((4))(0)=3y'''(0)=3*45=135`

Therefore, the simplified fourth order taylor polynomial is

`p_4(x)=10+5x+15/2 x^2+15/2 x^3+45/8 x^4`