Proof the following! $$a_0(x)\frac{d^2y}{dx^2}+a_1(x)\frac{dy}{dx}+a_2(x)y=0$$
A) Let $f_1$ & $f_2$ are two solutions to the above differential equation. Show that $f_1$ & $f_2$ are linearly independent on $a \leq x \leq b$ and $A_1$,$A_2$,$B_1$&$B_2$  are constants such that $A_1B_2-A_2B_1\neq 0$ then the solution $A_1f_1+A_2f_2$ & $B_1f_1+B_2f_2$ are also linearly independent on $a \leq x \leq b$ 
My work is down! 
I assume that the solution of the differential equation are linearly independent then we can write them as follows
$$A_1f_1+A_2f_2=0$$
$$\frac{f_1}{f_2}=-\frac{A_1}{A_2}$$
$$B_1f_1+B_2f_2=0$$
$$B_1 f'_1+B_2 f'_2=0$$
$$\frac{f'_1}{f'_2}=-\frac{B_1}{B_2}$$
Since the two solutions are linearly independent their Wronskian are not zero!
$$W[f_1(x),f_2(x)]=f_1f_2'-f_2f_1'\neq 0$$
$W(x)\neq 0$ therefore $ W'(x)\neq 0$
$$W(x)[f_1,f_2]=(-A_1)(-B_2)-(A_2)(B_1)$$
$$(-A_1)(-B_2)-(A_2)(B_1)\neq 0$$
What I do is very foolish. Can someone propose a proper way of doing things!
B) Let set ${f_1,f_2}$ be two solutions to the above differential equation and ${g_1,g_2}$ be another set then show that the wronskian is $W[f_1(x),f_2(x)]=cW[g_1(x),g_2(x)]$ such that $c\neq 0$
Since $f_1$ & $f_2$ are solutions, then 
$$a_0f_1''+a_1f_1'+a_2f_1+a_0f_2''+a_1f_2'+a_2f_2=0=0$$
$$a_0(f_1f_2''-f_2f_1'')+a_1(f_1f_2'-f_2f_1')+a_2(f_2f_1-f_1f_2)=0$$
$$a_0W'[f_1(x),f_2(x)]+a_1W[f_1(x),f_2(x)]=0$$
$$W'[f_1(x),f_2(x)]=-\frac{a_1}{a_0}W[f_1(x),f_2(x)]$$
$$W'[g_1(x),g_2(x)]=-\frac{a_1}{a_0}W[g_1(x),g_2(x)]$$
$$\int\frac{dW[g_1(x),g_2(x)]}{W[g_1(x),g_2(x)]}=\int \frac{dW[f_1(x),f_2(x)]}{W[f_1(x),f_2(x)]}$$
$$W[f_1(x),f_2(x)]=cW[g_1(x),g_2(x)]$$
Totally stuck!
This is bad. I can't even proceed from the question!
Hope someone help me in this question

This is pure linear algebra: If $f_1$ and $f_2$ are linearly independent vectors in some vector space and $ad-bc\ne0$ then $g_1:=a f_1+b f_2$ and  $g_2:=c f_1+d f_2$ are again linearly independent.
