Taylor series expansion doubt

Question

Employ taylor series method to obtain correct to four places of decimals, solution of the D.E with dy/dx =$x^2$ +$y^2$ with y=0 when x=0 for x=0.4

Solution: y'=$x^2$ +$y^2$

       y"=2x+2yy'  and so on

I dont understand how to evaluate (differentiate) y to get those y', y" and all. I am not able to figure out.

For e.g http://www.papersolutions.in/ans.php?sem=sem2&img=ans10&s=&m=may&y=13

I am not able to differentiate y, if possible kindly explain step by step execution.

Answer 1

You could also use the Picard iteration, $$ y_{n+1}(x)=0+\int_0^x(t^2+y_n(t)^2)\,dt =\frac13x^3+\int_0^xy_n(t)^2\,dt $$ Start with $y_0=0$ to get $y_1(x)=\frac13x^3$, insert to get $$ y_2(x)=\frac13x^3+\int_0^x\frac19t^6\,dt=\frac13x^3+\frac1{63}x^7 $$ which is precise up to $O(x^{11})$, and so on.

But already the term $\frac1{63}(0.4)^7\simeq 2.6·10^{-5}$ is below your error threshold.

Answer 2

You must use the rules for differentiating a composite function: $$ y''(x)=D(x^2+y(x)^2)=D(x^2)+D(y(x)^2)=2x + 2 y(x)D(y(x))=2x+2y(x)y'(x), $$ and so on.

Answer 3

You have \begin{align*} y' &= x^2 + y^2 \\ y'' &= 2x + 2 y y' \end{align*} so you have $$ y'' = 2x + 2y(x^2+y^2) \text{,} $$ i.e., you use your earlier equations to eliminate the $y'$, $y''$, et c. terms from the right-hand sides.