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Let $f: \mathbb{R} \rightarrow \mathbb{R}^n$ be $n$ times differentiable and let $G: \mathbb{R}^n \rightarrow \mathbb{R}$ be a function. Show that the $n^{th}$ order differential equation $f^{(n)}-G(f,f^{(1)},...,f^{(n-1)})=0$ can be reduced to a system of $n$ coupled first order differential equation

My attempt:

Let:

$$x_1 = f$$ $$x_2 = f^{(1)}$$ $$x_n = f^{(n-1)}$$

Then,

$$x_1'= x_2$$ $$x_2'= x_3$$ $$x_{n-1}'= x_n$$

Not sure where to go with this...

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    Show *what* about "the $n^{th}$ order differential equation $f^{(n)}-G(f,f^(1),...,f^{(n-1)})=0$"? You just have the equation and do not say what you want to show. – 2017-02-13

1 Answers 1

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You were nearly there. Let $x_1=f$, $x_2=f'$ and so on. Then we can rewrite the problem as $$x_1'=x_2$$ $$x_2'=x_3$$ $$\vdots$$ $$x_n'=G(x_1,x_2,\dots,x_n)$$ which is a system of coupled first order differential equations.

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    I hope I got your question right... it just seems you nearly had it. – 2017-02-14