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
ANSWERED
asked 2021-05-27

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
\(\displaystyle{y}^{{{\left({n}\right)}}}={F}{\left({x},{y},{y}',{y}{''},\ldots,{y}^{{{\left({n}-{1}\right)}}}\right)}?\)

Expert Answers (1)

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
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={z}_{{{1}}},{\frac{{{\left.{d}{z}\right.}_{{{1}}}}}{{{\left.{d}{x}\right.}}}}={z}_{{{2}}},{\frac{{{\left.{d}{z}\right.}_{{{2}}}}}{{{\left.{d}{x}\right.}}}}={z}_{{{3}}},\ldots{\frac{{{\left.{d}{z}\right.}_{{n}}-{2}}}{{{\left.{d}{x}\right.}}}}={z}_{{{n}-{1}}},{\frac{{{\left.{d}{z}\right.}_{{{1}}}}}{{{\left.{d}{x}\right.}}}}={F}{\left({x},{y},{z}_{{{1}}},{z}_{{{2}}},\ldots,{z}_{{{n}-{1}}}\right)}\)

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