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.`

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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)` .


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



Now once more.


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





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`

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