$$a\_0(x)\frac{d^2y}{dx^2}+a\_1(x)\frac{dy}{dx}+a\_2(x)y=0$$

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