Question

# Show that the second-order differential equation y″ = F(x, y, y′) can be reduced to a system of two first-order differential equations \frac{dy}{dx}=z

Differential equations

Show that the second-order differential equation $$y″ = F(x, y, y′)$$ can be reduced to a system of two first-order differential equations
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={z},{\frac{{{\left.{d}{z}\right.}}}{{{\left.{d}{x}\right.}}}}={F}{\left({x},{y},{z}\right)}.$$
Can something similar be done to the nth-order differential equation
$$\displaystyle{y}^{{{\left({n}\right)}}}={F}{\left({x},{y},{y}',{y}{''},\ldots,{y}^{{{\left({n}-{1}\right)}}}\right)}?$$

2021-05-28

Let $$\displaystyle{z}={y}'={\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}$$, then
$$\displaystyle{\frac{{{\left.{d}{z}\right.}}}{{{\left.{d}{x}\right.}}}}={z}'={y}{''}$$
The equation $$y''=F(x,y,y')$$ can be written as
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={z}$$
$$\displaystyle{\frac{{{\left.{d}{z}\right.}}}{{{\left.{d}{x}\right.}}}}={F}{\left({x},{y},{z}\right)}$$
Let $$\displaystyle{z}_{{{1}}}={y}'={\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}'}}}$$, then
$$\displaystyle{\frac{{{\left.{d}{z}\right.}_{{{1}}}}}{{{\left.{d}{x}\right.}}}}={z}'_{{{1}}}={y}{''}$$
Let $$\displaystyle{z}_{{{2}}}={z}'_{{{1}}}={y}{''}$$, then
$$\displaystyle{\frac{{{\left.{d}{z}\right.}_{{{2}}}}}{{{\left.{d}{x}\right.}}}}={z}'_{{{2}}}={y}{'''}$$
In the same way as before $$\displaystyle{z}_{{{n}-{1}}}={z}'_{{{n}-{2}}}={y}^{{{n}-{1}}}$$, then
$$\displaystyle{\frac{{{\left.{d}{z}\right.}_{{{n}-{1}}}}}{{{\left.{d}{x}\right.}}}}={z}'_{{{n}-{1}}}={y}^{{{n}}}$$
This gives the following system of differential equations of first order
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={z}_{{{1}}},{\frac{{{\left.{d}{z}\right.}_{{{1}}}}}{{{\left.{d}{x}\right.}}}}={z}_{{{2}}},{\frac{{{\left.{d}{z}\right.}_{{{2}}}}}{{{\left.{d}{x}\right.}}}}={z}_{{{3}}},\ldots{\frac{{{\left.{d}{z}\right.}_{{n}}-{2}}}{{{\left.{d}{x}\right.}}}}={z}_{{{n}-{1}}},{\frac{{{\left.{d}{z}\right.}_{{{1}}}}}{{{\left.{d}{x}\right.}}}}={F}{\left({x},{y},{z}_{{{1}}},{z}_{{{2}}},\ldots,{z}_{{{n}-{1}}}\right)}$$