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

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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John Koga

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.

###### Not exactly what you’re looking for?
Kindlein6h
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.
RizerMix

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.