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$$a_0(x)\frac{d^2y}{dx^2}+a_1(x)\frac{dy}{dx}+a_2(x)y=0$$

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

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You just need to show that $$W(g_1,g_2)=(A_1B_2−A_2B_1)W(f_1,f_2).$$ where $$ g_1=A_1f_1+A_2f_2,\\ g_2=B_1f_1+B_2f_2. $$ or $$ (g_1 \ \ g_2)=(f_1 \ \ f_2)\pmatrix{A_1&B_1\\A_2&B_2} $$


In A), already the first equation is wrong, as $g_1=A_1f_1+A_2f_2$ is not the zero function. Since $f_1,f_2$ are independent, $A_1f_1+A_2f_2=0$ is only possible for $A_1=A_2=0$.

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  • $\begingroup$ Can you please elaborate more on the first question? $\endgroup$ – Crazy Jul 10 '17 at 9:22
  • $\begingroup$ Thanks! I began to see everything after working them out. Last question to ask. I want to show that $f_1$ and $f_2$ both are linearly independent using the above method only right? Stating that W for g and f are essentially no zero for everything to hold? $\endgroup$ – Crazy Jul 10 '17 at 9:57
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    $\begingroup$ A single vector of function can not be linearly independent (or is trivially always so). Two non-zero vectors are linearly dependent if one is a scalar multiple of the other. You show that the zero sets of $W$ for $f_1,f_2$ and $g_1,g_2$ are the same $\endgroup$ – LutzL Jul 10 '17 at 10:18
  • $\begingroup$ Now I got it!!! $\endgroup$ – Crazy Jul 10 '17 at 10:21
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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.

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As for the first part, suppose $A_1f_1 + A_2 f_2$ and $B_1 f_1 + B_2 f_2$ are linearly dependent; the there exist constants $c_1$ and $c_2$, not both zero, such that

$c_1(A_1f_1 + A_2 f_2) + c_2(B_1 f_1 + B_2 f_2) = 0; \tag{1}$

re-arranging (1):

$(c_1A_1 + c_2 B_1)f_1 + (c_1 A_2 + c_2 B_2)f_2 = 0; \tag{2}$

since $f_1$ and $f_2$ are linearly independent, we must have

$c_1 A_1 + c_2 B_1 = 0 \tag{3}$

$c_1 A_2 + c_2 B_2 = 0; \tag{4}$

(3) and (4) may be written as

$\begin{bmatrix} A_1 & B_1 \\ A_2 & B_2 \end{bmatrix} \begin{pmatrix} c_1 \\ c_2 \end{pmatrix} = 0; \tag{5}$

since $A_1 B_2 - B_1 A_2 \ne 0$, the $2 \times 2$ matrix on the right of (5) is nonsingular and thus we conclude that

$\begin{pmatrix} c_1 \\ c_2 \end{pmatrix} = 0, \tag{6}$

that is,

$c_1 = c_2 = 0, \tag{7}$

in contradiction to the assumption that $c_1$ and $c_2$ are not both $0$. This contradiction shows that $A_1f_1 + A_2 f_2$ and $B_1 f_1 + B_2 f_2$ are linearly independent.

As for part (B), I am going to assume $a_0(x) \ne 0$ for $x \in [a, b]$ as is explained in my answer to Verification for the solution following differential equation!, which adresses the same differential equation. In that post, it is shown that the Wronskian $W(f_1, f_2)(x)$ satisfies

$W(f_1, f_2)(x) = \exp(- \displaystyle \int_a^x\dfrac{a_1(s)}{a_0(s)}ds) W(f_1, f_2)(a); \tag{12}$

likewise, thecWronskian for $g_1$, $g_2$ also obeys

$W(g_1, g_2)(x) = \exp(- \displaystyle \int_a^x\dfrac{a_1(s)}{a_0(s)}ds) W(g_1, g_2)(a); \tag{13}$

since $f_1$, $f_2$ are linearly independent, as are $g_1$, $g_2$, we have

$W(f_1, f_2)(a) \ne 0 \ne W(g_1, g_2)(a), \tag{14}$

and thus there exists $c \ne 0$ such that

$W(f_1, f_2)(a) = c W(g_1, g_2); \tag{15}$

thus, using (12), (13) and (15),

$W(f_1, f_2)(x) = \exp(- \displaystyle \int_a^x\dfrac{a_1(s)}{a_0(s)}ds) W(f_1, f_2)(a) = \exp(- \displaystyle \int_a^x\dfrac{a_1(s)}{a_0(s)}ds) cW(g_1, g_2)(a) = c \exp(- \displaystyle \int_a^x\dfrac{a_1(s)}{a_0(s)}ds) W(g_1, g_2)(a); = cW(g_1, g_2)(x), \tag{16}$

as required.

I like this argument because it doesn't explicitly use the fact that $g_1$, $g_2$ are linear combinations of $f_1$, $f_2$.

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