Find general solution of Differential equation if you know three solutions. I tried to solve this problem, however I have a question about the Wronskian. Three particular solutions are 1, $x$, ${x^2}$. I know, that general solution can be written as $C_1{(x^2-1)} +C_2{(x-1) + 1}$, because difference of two particular is general. However, there is a theorem that says if Wronskian equals zero in one point, then the functions are linearly dependent. Proof that ODE solutions with Wronskian identically zero are linearly dependent. And I'm stuck here, ${W[x^2-1,x-1]}$ is zero when ${x=1}$, so this functions can't form a general solution, as they linearly dependent. But in the textbook $C_1{(x^2-1)} +C_2{(x-1) + 1}$ is a solution, where is a mistake in my reasoning? I totally understand that ${1,x,x^2}$ are linearly independent, I just want to understand, why this theorem doesn't work and where is my mistake. I can also give a proof of the theorem, as short, as I can: let's consider ${y(x)}$ is a solution to a differential equation with general solution $C_1{(x^2-1)} +C_2{(x-1)}$, then ${y(x) = C_1{(x^2-1)} +C_2{(x-1)}.}$ Let ${C_1 = 1, C_2=-2}$. ${y(1) =0}$ and ${y'(1)=0}$ and there exists only one solution by existence and uniqueness theorem. Therefore ${y(x)=0}$ and $C_1{(x^2-1)} +C_2{(x-1)}$ linearly dependent.
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You can verify that $y(x)=C_1(x^2-1)+C_2(x-1)$ is the general solution to the second order linear ODE $$ y''-\frac{2}{x-1}y'+\frac{2}{(x-1)^2}y=0. \tag{1} $$ The proof of the theorem you mention assumes that the coefficients $p_0(x), p_1(x),\ldots,p_{n-1}(x)$ of the linear differential equation $$ y^{(n)}+p_{n-1}(x)y^{(n-1)}+\ldots+p_1(x)y'+p_0(x)y=0 \tag{2} $$ are continuous. In Eq. $(1)$, however, $p_0(x)=\frac{2}{(x-1)^2}$ and $p_1(x)=-\frac{2}{x-1}$ are not continuous at $x=1$, so the vanishing of the Wronskian $W[x^2-1,x-1]$ at $x=1$ does not imply that $x^2-1$ and $x-1$ are linear dependent functions.
Gonçalo
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