So here's the problem I'm having trouble with...
The initial value problem:
$$ x'(t) = \cos(x+t^2) $$ with $x(-5) = -1$ on the interval $[-5, 3]$.
I have no idea how to star this problem off and need some advice.
I'm looking at a practice example and this is how they did it.
$$ x'(t) = (x+t)^2 $$ and the second derivative is $$ x''(t) = 2(x+t) (x'(t) + 1) $$
I don't understand why they put x'(t) into the second derivative...
On
The problem asks to solve the ordinary differential equation
$$x'(t)=cos(x+t^2)$$
in the interval I=[-5, 3] with the the initial value $x(-5)=-1$ by a Taylor series
$$x(t) = \sum_{i=0}^{\infty} a_{_i} (t-t_0)^i$$
and to compute the coefficients $a_i, i = 0,1,2$.
The differential equation satisfies a global Lipschitz condition. Hence a unique solution $x(t), t \in I$, exists.
We choose $t_0 = -5$, then $a_0 = -1$. Moreover
$$cos \ z = 1 - (1/2) z^2 + \mathscr O(4)$$
$$t^2 = (t+5)^2 -10(t+5) +25$$
$$x(t) = -1 + a_1(t+5) + \mathscr O(2)$$
$$x'(t) = a_1 +2a_2(t+5) + \mathscr O(2).$$
Comparing coefficients for $x'(t)=cos(x+t^2)$ gives
$$x'(t) = 1 -(1/2)[-1 + a_1(t+5) -10(t+5) +25]^2 + \mathscr O(2)$$
$$x'(t) = 1 -(1/2)[24 + (a_1-10)(t+5) ]^2 + \mathscr O(2)$$
$$x'(t) = 1 -(1/2)[24^2 + 48(a_1-10)(t+5) + \mathscr O(2)]$$
$$a_1 +2a_2(t+5) + \mathscr O(2) = -287 -24(a_1-10)(t+5) + \mathscr O(2).$$
Comparing coefficients on both sides gives
$$a_1 = -287, a_2 = 3564.$$
We can write
\begin{align*} x(t+h) &= x(t) + hx^{\prime}(t) + \frac{h^{2}}{2}x^{\prime\prime}(t) + \mathcal{O}(h^{3})\\ & = x(t) + h\cos(x+t^{2}) + \frac{h^{2}}{2}x^{\prime\prime}(t) + \mathcal{O}(h^{3}) \end{align*}
Since \begin{align*} x^{\prime\prime} &= [x^{\prime}]^{\prime}\\ &= [\cos(x+t^{2})]^{\prime}\\ &= -2t\sin(x+t^{2})-\sin(x+t^{2})x^{\prime}\\ &= -2t\sin(x+t^{2}) - \sin(x+t^{2})\cos(x+t^{2})\end{align*}
We get that
$$x(t+h) = x(t) + h\cos(x+t^{2}) + \frac{h^{2}}{2}\left(-2t\sin(x+t^{2}) - \sin(x+t^{2})\cos(x+t^{2})\right) + \mathcal{O}(h^{3})$$