# Wrong reasoning for uniqueness of solution of ODE?

Sometimes I have seen this argument to prove that a differential equation has an unique solution, but I think it's wrong.

Suppose the differential equation: $$\mathscr{D}[y(t)]=f(x)$$ where $\mathscr{D}$ is an arbitrary differential operator of order $N$. With initial conditions: $$y(0)=K_0,Dy(0)=K_1\dots,D^{N-1}y(0)=K_{N-1}$$ has two solutions $y_1,y_2$. Then $w=y_1-y_2$ is a solution of the initial value problem: $$\mathscr{D}[y(t)]=0$$ With initial conditions: $$y(0)=0, Dy(0)=0\dots,D^{N-1}y(0)=0$$ As this second problem has solution $w=0$ then $y_1=y_2$.

I think this argument is wrong because the second system could have more solutions. Then maybe it has one solution $w_1=0$, but another $w_2\neq 0$. It looks if we have just passed the issue from one initial value problem to another.

Anybody could help me to make clear this doubt?

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The reasoning is correct: there is an existence and uniquiness theorem for the homogeneous initial value problem you are considering. This is the well known Picard Lindelof theorm applied to the equivalent system of $N$ first order Cauchy's problems. Being $0$ obviously Lipschitz continuous the theorem applies, therefore there exist only one solution to the problem $$\mathcal{D}[y(t)] = 0, \qquad y(0) = 0, \dots ,\ D^{N-1}y(0) = 0.$$