Let be $f:\mathbb{R}^2\to\mathbb{R}$ twice differentiable, such that $$f(1,0)=2,\quad \nabla f(1,0)= \begin{bmatrix} 0 \\ 1 \\ \end{bmatrix} \quad \nabla^2f(1,0)=\begin{bmatrix} 1 & -1 \\ -1 & 0 \\ \end{bmatrix} $$ and consider the initial value problem
$\frac{dy}{dx}=f(x,y), \quad y(1)=0,\quad x \in \mathbb R.$
Determine the Taylor approximation of order 3 for the solution $y(x)$ in a neighborhood of $x(1)$
I have doubts with the solutions, I do not know which is fine. Please can you help me
SOL:
I did these two ways, I do not know which is fine.
Sol 1
Taylor's approximation of order 3 for the solution y(x) in a neighborhood of x(1)
$y(x)\approx y(1)+\frac{dy(1)}{dx}(x-1)+\frac{1}{2}\frac{dy^2(1)}{dx^2}(x-1)^2+\frac{1}{6}\frac{dy^3(1)}{dx^3}(x-1)^3$
of hypothesis we have
$y(1)=0$ $\quad \quad \quad \quad \quad \quad \quad \quad$... (1)
$\frac{dy}{dx}=f(x,y) \quad \Rightarrow \quad \frac{dy(1)}{dx}=f(1,0)=2 $ $\quad \quad \quad \quad \quad \quad \quad \quad$... (2)
$\frac{dy^2}{dx^2}=\frac{d}{dx}(\frac{dy}{dx})= \frac{\partial}{\partial x}(f(x,y))\quad \Rightarrow \quad \frac{dy^2}{dx^2}=\frac{\partial}{\partial x}(f(x,y))=0$ (of $\nabla f(1,0)$ ) $\quad\quad $ (3)
$\frac{d}{dx}(\frac{dy^2}{dx^2})=\frac{d}{dx}( \frac{\partial}{\partial x}(f(x,y)))\quad \Rightarrow \quad \frac{dy^3}{dx^3}=\frac{\partial^2}{\partial x^2}(f(x,y))=1$ (of$ \nabla ^2 f(1,0)$ ) $\quad\quad $ (4)
substituting in the Taylor's approximation of order 3, we get
$y(x)=2(x-1)+\frac{1}{6}(x-1)^3$
Sol 2.
of hypothesis we have
$y(1)=0$ $\quad \quad \quad \quad \quad \quad \quad \quad$... (1´)
$\frac{dy}{dx}=f(x,y) \quad \Rightarrow \quad \frac{dy(1)}{dx}=f(1,0)=2 $ $\quad \quad \quad \quad \quad \quad \quad \quad$... (2´)
$\frac{dy^2}{dx^2}=\frac{d}{dx}(\frac{dy}{dx})= \frac{d}{dx}(f(x,y))= \frac{\partial}{\partial x}f(x.y)\frac{dx}{dx}+\frac{\partial}{\partial y}f(x.y)\frac{dy}{dx}$
$\quad \Rightarrow \quad \frac{dy^2}{dx^2}=0.1+1.2=2$ (of $\nabla f(1,0)$ ) $\quad\quad $ (3´)
$\frac{d}{dx}(\frac{dy^2}{dx^2})=\frac{d}{dx}(\frac{\partial}{\partial x}f(x.y)+\frac{\partial}{\partial y}f(x.y)\frac{dy}{dx})=\frac{d}{dx}(\frac{\partial}{\partial x}f(x,y)) + \frac{d}{dx}(\frac{\partial}{\partial y}f(x.y)f(x,y))$
applying chain rule we get
$\frac{dy^3}{dx^3}= \frac{\partial^2}{\partial x^2}(f(x,y))+\frac{\partial}{\partial y}f(x,y).f(x,y)+\frac{\partial^2}{\partial x \partial y}(f(x,y)).f(x,y)+\frac{\partial }{\partial y}(f(x,y)).\frac{\partial }{\partial x}(f(x,y))$
$\quad \Rightarrow \quad \frac{dy^3}{dx^3}=1+1.2+(-1).2+1.0=1$
(of$ \nabla ^2 f(1,0)$ ) $\quad\quad $ (4´)
substituting in the Taylor's approximation of order 3, we get
$y(x)=2(x-1)+(x-1)^2+\frac{1}{6}(x-1)^3$
I have doubts with the solutions, I do not know which is fine. Please can you help me
The second variant looks good.
In the first variant you wrongly change a total derivative into a partial derivative, while in the second variant that is correctly handled.