linearly independent solution to second order ODE.

If $y(t)$ is a known solution them substituting $z(t) = \lambda(t)y(t)$ into the DE we have

$$ y\ddot\lambda + (a y+2\dot y)\dot\lambda + \ddot y+a \dot y+b y = 0 $$

but

$$ \ddot y+a\dot y+b y = 0 $$

then

$$ y\ddot\lambda + (a y +2\dot y)\dot\lambda = 0 $$

so making $\dot\lambda = v$ we have

$$ y\dot v + (a y+2\dot y)v = 0 $$

now solving for $v$ after integrating we can obtain $z = \lambda y$ as a new independent solution.

First of all, we say that $y$ and $z = yv$ are linearly independent iff:

$$Ay + Bz = 0 ~\forall t \iff A=B=0.$$

Notice that:

$$Ay + Byv = y(A+Bv).$$

If $v$ is constant for all $t$, then $$A=-Bv \Rightarrow Ay + Byv = 0.$$

Therefore, we need that $v$ is not constant over time in order to have that $y$ and $z$ are linearly independent.

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Since $y(t)$ is a solution of $\ddot{x}+a(t)\dot{x}+b(t)x=0$, then:

$$\ddot{y}+a(t)\dot{y}+b(t)y=0.$$

Let's consider $z(t) = y(t)v(t)$. If $z(t)$ is a solution of $\ddot{x}+a(t)\dot{x}+b(t)x=0$, then:

$$\ddot{z}+a(t)\dot{z}+b(t)z=0 \Rightarrow \\ \ddot{(yv)}+a(t)\dot{(yv)}+b(t)yv=0 \Rightarrow \\ \ddot{y}v+2\dot{y}\dot{v}+y\ddot{v}+a(t)(\dot{y}v + y\dot{v})+b(t)yv=0 \Rightarrow \\ v(\ddot{y}+ a(t)\dot{y} + b(t)y)+ 2\dot{y}\dot{v}+y\ddot{v}+a(t)y\dot{v}=0 \Rightarrow \\ v \cdot 0+ 2\dot{y}\dot{v}+y\ddot{v}+a(t)y\dot{v}=0 \Rightarrow \\ y\ddot{v}+ (2\dot{y}+a(t)y)\dot{v}=0.$$

Now, let's introduce $w = \dot{v}$. The previous equation can be rewritten as:

$$\begin{cases} \dot{v} & = & w\\ y\dot{w} & = & -(2\dot{y}+a(t)y)w \end{cases}$$

This means that the non constant function $v$ satisfies the previous system of ODEs.