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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,...


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`


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