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
http://mathworld.wolfram.com/TaylorSeries.html