I have got a problem from Wronskian..I am a first reader of Differential Equation.
Can anyone please help me to solve this problem?
Attempt: I know that if $n$ number of $n-1$ differentiable function are dependent on $I$ then their Wronskian will be identically zero on $I$. But the converse of the theorem not true.
So by this theorem I can not do anything. If I had found a $x$ for which wronskian at the point $x$ is not zero , then we could have said the solutions are independent. But there is no such $x$.
Can anyone please correct me if I have misunderstood anywhere?
Since $u_{1}$ and $u_{2}$ are two solutions of the given differential equation, so
$ \frac{d^2 u_1}{dx^2}+P(x)\frac{du_1}{dx}+Q(x)u_1 = 0$ . . . . . (1)
$ \frac{d^2 u_2}{dx^2}+P(x)\frac{du_2}{dx}+Q(x)u_2 = 0$ . . . . . (2)
Multiplying (2) by $u_1$ and (1) by $u_2$ and then subtracting we get,
$(u_2'' u_1 - u_1'' u_2)+P(x) (u_2' u_1 - u_1' u_2)+Q(x) (u_2 u_1 - u_1 u_2)=0$ where $u''=\frac{d^2 u}{dx^2}$ and $u'=\frac{du}{dx}$
$\implies \frac{dW}{dx}+P(x) W= 0$, where $W=\begin{vmatrix} u_1 & u_2 \\ u_1' & u_2' \end{vmatrix}$
$\implies W=ce^{-\int {P(x)} dx}$ , where $c$ is a constant of integration. . . . . .(3)
For $x=0$, $W=\begin{vmatrix} u_1(0) & u_2(0) \\ u_1'(0) & u_2'(0) \end{vmatrix}=\begin{vmatrix} 0 & 0 \\ u_1'(0) & u_2'(0) \end{vmatrix}=0$, since $u_1(0)=u_2(0)=0$(given)
Now $W=0$ for $x=0$ and hence from equation (3) we get, $c=0$
Therefore $W(x)=0$ remains same in the nbd. of $x=0$
Since Wronskian is zero, so $u_1(x), u_2(x)$ are linearly dependent. [Option 2 is correct]
Since $[-1,1]$ is a nbd. of $x=0$, so $W(x)=0$, for $x\in[-1,1]$. [Option 3 is correct]