# Taylor's series method to solve first order first degree ODE

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Shrivatsa Joshi on 27 May 2023
Commented: Shrivatsa Joshi on 30 May 2023
Can someone help with how to solve first order first degree ODE using Taylor's series method? I am able to find derivatives but unable to substitute the values at given initial condition.
I have typed the code as below,
clear
clc
syms x y(x)
x0 = 0;
y0 = 1;
y1 = x^2*y-1;
y2 = diff(y1);
y3 = diff(y2);
y4 = diff(y3);
y10 = subs(y1,x,0);
y20 = subs(y2,x,0);
y30 = subs(y3,x,0);
y40 = subs(y4,x,0);
%Taylor's Method
y = y0 + (x-x0)*y10 +(((x-x0)^2)/2)*y20 + (((x-x0)^3)/6)*y30 + (((x-x0)^4)/24)*y40
And the output is as below. substitution at x=0 is not working. Kindly help.

VBBV on 27 May 2023
Edited: VBBV on 27 May 2023
Use taylor function
clear
clc
syms x y(x)
x0 = 0;
y0 = 1;
y1 = x^2*y-1;
y2 = diff(y1);
y3 = diff(y2);
y4 = diff(y3);
y10 = subs(y1,x,0);
y20 = subs(y2,x,0);
y30 = subs(y3,x,0);
y40 = subs(y4,x,0);
%Taylor's Method
% y = y0 + (x-x0)*y10 +(((x-x0)^2)/2)*y20 + (((x-x0)^3)/6)*y30 + (((x-x0)^4)/24)*y40
% use taylor
y = y0+taylor(x^2*y-1,x,'ExpansionPoint',0)
y(x) = Shrivatsa Joshi on 30 May 2023
If you consider taylor function after first term (y0), the coefficient of y'(0) will be taken as 1, which is wrong. The coefficient of y'(0) should be x.
And moreover taylor function can be used on known function in MATLAB to expand. Not to find the unknown function.