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Solve: $$y_1' =y_2 y_3 , y_2 ' = -y_1 y_3 , y_3 '= 2$$

for $(y_1(0),y_2(0),y_3 (0)= (0,1,0) \in \mathbb{R}^3)$

First I can see fastly that: $y_3 = 2x (+K)$ from that I can say solution is: $$(\sin(x^2),\cos(x^2),2x)$$

Is there a fast way to find the approximation by Picard iteration?

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1 Answer 1

Having found $y_3$, you are left with the system $y_1'=2xy_2$, $y_2'=-2xy_1$. Seeing $2x$ as a factor in both equations, it is natural to kill it with a chain: $u=x^2$ yields $$\frac{dy_i}{dx} = \frac{dy_i}{du}\frac{du}{dx} = 2x \frac{dy_i}{du}$$ hence $\frac{dy_1}{du}=y_2$ and $\frac{dy_2}{du}=-y_1$. For the constant-coefficient system we know what to do... this is probably how you found the solution already.

The Picard iteration would work too, but more slowly. Set it up as $$y_1(x) = \int_0^x 2ty_2(t)\,dt, \quad y_2(x) =1- \int_0^x 2ty_1(t)\,dt$$ and run iteration starting with $(y_1,y_2)\equiv (0,1)$. I omit indices for the steps of iteration: $$\begin{align}y_1(x) = x^2, \quad &y_2(x) =1 \\ y_1(x) = x^2, \quad &y_2(x) =1-\frac{x^4}{2} \\ y_1(x) = x^2-\frac{x^6}{6}, \quad &y_2(x) =1-\frac{x^4}{2} \\ y_1(x) = x^2-\frac{x^6}{6}, \quad &y_2(x) =1-\frac{x^4}{2}+\frac{x^8}{24} \end{align}$$ and so on. It's pretty neat, actually.

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