I'm studying the steps to obtain the general solution of a second order Linear ODE with constant coefficients, but I haven't understood a justification.

$$y''+ay'+by=f$$ Let's look for a solution $y=\gamma_1 y_1 +\gamma_2 y_2$ where: $\gamma_1, \gamma_2 \in C^1(R)$ and $y_1, y_2$ solutions of the associated homogeneous equation.

We arrive at $$\begin {cases} \gamma_1 ' y_1+ \gamma_2 ' y_2=0\\ \gamma_1'y_1'+\gamma_2'y_2'=f(x)\end{cases} $$

The system has only one solution if $$ Det\begin{pmatrix} y_1(x) & y_2(x)\\ y_1(x)' & y_2'(x)\end {pmatrix} \ne 0$$

And (this is the step that I haven't understood) it is $\ne0$ for all $ x\in R$, because of $y_1$ and $ y_2$ are linearly independent.

But I know that if $y_1$ and $y_2$ are linearly independent, $\exists$ $x_0 \in R$ such as the Wronskian $\det W(x_0)\ne0$.

I deduce that $\exists x_0$ such as I can have only one solution of the system. But why the book says that $\forall x \in R \; \det W(x) \ne 0$?

Many thanks


1 Answer 1



I know that if $y_1$ and $y_2$ are linearly independent, $\exists x_0 \in \mathbb{R}$ such as the wronskian ${\rm det }\; W(x_0) \neq 0$.


But why the book says that $W(x)\ne 0 \; \forall x \in R?$

This is also a property of Wronskian: it can't be zero only at one point. If there is a point at which Wronskian is zero, you can deduce that the set of solutions is linear dependent at whole $\mathbb{R}$. An elementary proof for this property can be found in almost any book about ODEs (roughly speaking,proof relies on the fact that linear dependence at one point allows to construct second solution to some IVP which contradicts uniqueness and existence theorem). So, here is a dichotomy: $\det W \equiv 0$ or it is non-zero everywhere on $\mathbb{R}$.

So, this property allows you to solve system of linear equations at each point $x \in \mathbb{R}$.


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