Showing given solution solves ODE?

Let $a(x),b(x)$ two continuous functions and $v=\begin{pmatrix} v_1 \\ v_2\end{pmatrix}$ a solution to the ODE $$y'=\begin{pmatrix}0 & 1 \\ -b(x) & a(x)\end{pmatrix}y$$ Additionally $v_1 \neq0$ and $w$ is a non-constant solution to the ODE $$y'=\begin{pmatrix}0 & 1 \\ 0 & -2\frac{v_1'}{v_1} + a(x)\end{pmatrix}y$$ Proof that $u=\begin{pmatrix} (v_1 w_1) \\ (v_1w_1)'\end{pmatrix}$ is a second solution to the first ODE.

As far as I know it would suffices, if we plug in the solution and see that it'll work , but this doesn't work for me because

$u'=\begin{pmatrix} (v_1 w_1)' \\ (v_1w_1)''\end{pmatrix}=$ $\begin{pmatrix}0 & 1 \\ -b(x) & a(x)\end{pmatrix}\begin{pmatrix} (v_1 w_1) \\ (v_1w_1)'\end{pmatrix}$

Now I'm stuck on $(v_1w_1)''=-b(x)(v_1w_1)+a(x)(v_1w_1)'$

Maybe for some $a(t),b(t)$ this is a true statement, but I don't know which ones.

This is the info you currently have: \begin{align*} \begin{pmatrix} v_1 ' \\ v_2' \end{pmatrix} &= \begin{pmatrix} 0 & 1 \\ -b(x) & a(x) \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix}\\[0.3cm] \begin{pmatrix} w_1 ' \\ w_2' \end{pmatrix} &= \begin{pmatrix} 0 & 1 \\ 0 & -2\frac{v_1'}{v_1} + a(x) \end{pmatrix} \begin{pmatrix} w_1 \\ w_2 \end{pmatrix} \end{align*}

It can be rewritten as follows: \begin{align*} v_1' &= v_2\\ v_2' &= -b(x) v_1 + a(x) v_2 \tag{1}\\ w_1' &= w_2\\ w_2' &= -2\frac{v_1'}{v_1} w_2 + a(x)w_2 \tag{2} \end{align*} Equation $(2)$ gives this: $$v_1 w_2' = -2 v_1' w_2 + a(x)v_1w_2$$ And since $w_1' = w_2$, then $w_1'' = w_2'$ and we have: $$v_1 w_1'' = -2 v_1' w_1' + a(x)v_1w_1' \tag{3}$$

You want to show this: $$(v_1 w_1)'' = -b(x) (v_1w_1) + a(x) (v_1w_1)'$$

Expand on the LHS to get this: $$v_1''w_1 + 2v_1'w_1' + v_1w_1'' = -b(x) (v_1w_1) + a(x) (v_1w_1)' \tag{4}$$

What we're going to do is manipulate the LHS of Equation $(4)$ to make it look like the RHS. First substitute Equation $(3)$ into the LHS of Equation $(4)$ above and simplify to get: $$v_1''w_1 + a(x) v_1 w_1' = -b(x) (v_1w_1) + a(x) (v_1w_1)' \tag{5}$$

Since $v_1' = v_2$, then $v_1'' = v_2'$ and from Equation $(1)$ above we have: $$v_1'' = -b(x) v_1 + a(x) v_1'$$

Now multiply both sides by $w_1$ to get: $$v_1''w_1 = -b(x) v_1w_1 + a(x) v_1'w_1 \tag{6}$$

Substitute Equation $(6)$ into the LHS of Equation $(5)$ to get: $$-b(x) v_1w_1 + a(x) v_1'w_1 + a(x) v_1 w_1' = -b(x) (v_1w_1) + a(x) (v_1w_1)' \tag{7}$$ Notice we have the $b(x)$ term that we want. Next: \begin{align*} a(x)v_1'w_1 + a(x) v_1w_1' &= a(x) \left(v_1'w_1 + v_1w_1'\right)\\ &= a(x) (v_1w_1)' \end{align*} The last equality above follows from the product rule. Finally, substitute that into Equation $(7)$ to get the desired result: $$-b(x) v_1w_1 + a(x) (v_1w_1)' = -b(x) (v_1w_1) + a(x) (v_1w_1)'$$