linearly independent solution to second order ODE. 
Let $y(t)$ be a nontrivial solution for the second order differential equation
$\ddot{x}+a(t)\dot{x}+b(t)x=0$
to determine a solution that is linearly independent from $y$ we set $z(t)=y(t)v(t)$.
Show that this leads to a first order differential equation for $\dot{v}=w$

What do they even mean with linearly independent from $y$ ? Is it meant in the sense that the solution shouldn't be just different by a constant from $y$ like in linear algebra ?
Also how would I go on showing that this leads to a solution of first order ? I'm kind of lacking any idea where to start.
 A: 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.
A: 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.

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