Find the approximate solution of the initial value problem presented below in te

Question

Find the approximate solution of the initial value problem presented below in terms of a Taylor polynomial of degree 4.

y?+x^2 ?y^2 =0, y(0)=1

Explanation / Answer

y' x^2-y^2=0 y(0)=1 y'= y^2/x^2 y'=f(x,y) y'= y^2/x^2==> y'' = 2yy' / x^2 = 2y(y^2/x^2) /x^2 ==> y'' = 2y^3 / x^4 y'''= 6 y^2 y' / x^4 = 6y^2(y^2 /x^2) x^4 = 6y^4 / x^6 y'''' = 24 y^3 y' / x^6= 24 y^3 (y^2 /x^2 ) / x^6 = 24y^5 / x^8 y(x_{n+1}) = y(x_[n}) + h y'(x_{n}) + (h^2) y''(x_{n}) /(2!) +(h^3) y'''(x_{n}) /(3!) + (h^4) y''''(x_{n}) /(4!) y(x_{n+1}) = y(x_[n}) + h ( ( y(x_{n}) /(x_{n}))^2 + (h^2) ( 2 ( y(x_{n}) )^3 / (2! (x_{n})^4 ) +(h^3) ( 6 ( y(x_{n}) )^4 / (3! (x_{n})^6 ) + (h^4) ( 24 ( y(x_{n}) )^5 / (4! (x_{n})^8 ) where h= stepsize

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