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

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.

Answer 1

[VBBV]

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]

y(0), y'(0),y''(0) and y'''(0) all have to be substituted by the program itself. Which is not happening here.

Also the coefficients are wrongly being considered.

[VBBV]

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)

y10(x) = 

y20 = subs(y2,x,0)

y20(x) = 

0

y30 = subs(y3,x,0)

y30(x) = 

y40 = subs(y4,x,0)

y40(x) = 

%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) = 

y = subs(y,x,0)

y(x) = 

0

Can you tell why coefficients are not considered correctly ?

[Shrivatsa Joshi]

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.