Let y(t) be a nontrivial solution for the second order

Painevg 2021-12-18 Answered
Let \(\displaystyle{y}{\left({t}\right)}\) be a nontrivial solution for the second order differential equation
\(\displaystyle\ddot{{{x}}}+{a}{\left({t}\right)}\dot{{{x}}}+{b}{\left({t}\right)}{x}={0}\)
to determine a solution that is linearly independent from y we set \(\displaystyle{z}{\left({t}\right)}={y}{\left({t}\right)}{v}{\left({t}\right)}\)
Show that this leads to a first order differential equation for \(\displaystyle\dot{{{v}}}={w}\)

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Expert Answer

John Koga
Answered 2021-12-19 Author has 4212 answers

First of all, we say that y and \(\displaystyle{z}={y}{v}\) are linearly independent if:
\(\displaystyle{A}{y}+{B}{z}={0}\forall\Rightarrow{A}={B}={0}\)
Notice that:
Notice that:
\(\displaystyle{A}{y}+{B}{y}={y}{\left({A}+{B}{v}\right)}\)
If v is constant for all t, then
\(\displaystyle{A}=-{B}{v}\Rightarrow{A}{y}+{B}{y}{v}={0}\)
Therefore, we need that v is not constant over time in order to have that y and z are linearly independent.
Since \(\displaystyle{y}{\left({t}\right)}\ {i}{s}\ {a}\ {s}{o}{l}{u}{t}{i}{o}{n}\ {o}{f}\ \ddot{{{x}}}+{a}{\left({t}\right)}\dot{{{x}}}+{b}{\left({t}\right)}{x}={0}\), then:
\(\displaystyle\ddot{{{y}}}+{a}{\left({t}\right)}\dot{{{y}}}+{b}{\left({t}\right)}{y}={0}\)
Let's consider \(\displaystyle{z}{\left({t}\right)}={y}{\left({t}\right)}{v}{\left({t}\right)}.{I}{f}\ {z}{\left({t}\right)}\ {i}{s}\ {a}\ {s}{o}{l}{u}{t}{i}{o}{n}\ {o}{f}\ \ddot{{{x}}}+{a}{\left({t}\right)}\dot{{{x}}}+{b}{\left({t}\right)}{x}={0}\) then:
\(\displaystyle\ddot{{{z}}}+{a}{\left({t}\right)}\dot{{{z}}}+{b}{\left({t}\right)}{z}={0}\Rightarrow\)
\(\displaystyle{\left(\ddot{{{y}}}{v}\right)}+{a}{\left({t}\right)}{\left(\dot{{{y}}}{v}\right)}+{b}{\left({t}\right)}{y}{v}={0}\Rightarrow\)
\(\displaystyle\ddot{{{y}}}{v}+{2}\dot{{{y}}}\dot{{{v}}}+{y}\ddot{{{v}}}+{a}{\left({t}\right)}{\left(\dot{{{y}}}{v}+{y}\dot{{{v}}}\right)}+{b}{\left({t}\right)}{y}{v}={0}\Rightarrow\)
\(\displaystyle{v}{\left(\ddot{{{y}}}+{a}{\left({t}\right)}\dot{{{y}}}+{b}{\left({t}\right)}{y}\right)}+{2}\dot{{{y}}}\dot{{{v}}}+{a}{\left({t}\right)}{y}\dot{{{v}}}={0}\)
\(\displaystyle{v}\cdot{0}+{2}\dot{{{y}}}\dot{{{v}}}+{y}\ddot{{{v}}}+{a}{\left({t}\right)}{y}\dot{{{v}}}={0}\)
\(\displaystyle{y}\ddot{{{v}}}+{\left({2}\dot{{{y}}}+{a}{\left({t}\right)}{y}\right)}\dot{{{v}}}={0}\)
Now let's introduce \(\displaystyle{w}=\dot{{{v}}}\). The previous equation can be rewritten as:
\(\displaystyle{\left\lbrace\begin{array}{c} \dot{{{v}}}={w}\\{y}\dot{{{w}}}=-{\left({2}\dot{{{y}}}+{a}{\left({t}\right)}{y}\right)}{w}\end{array}\right.}\)
This means that the non constant function v satisfies the previous system of ODEs.

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Kindlein6h
Answered 2021-12-20 Author has 3362 answers
If \(\displaystyle{y}{\left({t}\right)}\) is a known solution them substituting \(\displaystyle{z}{\left({t}\right)}=\lambda{\left({t}\right)}{y}{\left({t}\right)}\) into the DE we have
\(\displaystyle{y}\ddot{{\lambda}}+{\left({a}{y}+{2}\dot{{{y}}}\right)}\dot{{\lambda}}+\ddot{{{y}}}+{a}\dot{{{y}}}+{b}{y}={0}\)
but
\(\displaystyle\ddot{{{y}}}+{a}\dot{{{y}}}+{b}{y}={0}\)
then \(\displaystyle{y}\ddot{{\lambda}}+{\left({a}{y}+{2}\dot{{{y}}}\right)}\lambda{\left\lbrace{y}\right\rbrace}={0}\)
so making \(\displaystyle\lambda\cdot={v}\) we have
\(\displaystyle{y}\dot{{{v}}}+{\left({a}{y}+{2}\dot{{{y}}}\right)}{v}={0}\)
now solving for v after integrating we can obtain \(\displaystyle{z}=\lambda{y}\) as a new independent solution.
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RizerMix
Answered 2021-12-29 Author has 9350 answers

The instructions are given: set z=yv and obtain an equation in \(\dot{v}\), rewritten w
The substitution give
\(\ddot{y}v+2\dot{y}w+\dot{w}+a\dot{y}v+ayw+byv=0\)
As y is a solution, a simplification occurs, and
\(2\dot{y}w+\dot{w}+ayw=0\)
In the latter, t and y are known.

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