Question about second order differential equation

[Telemachus]

Homework Statement

I have this problem, which says: If the graph of one solution for the equation y''+P(x)y'+Q(x)y=0 is tangent to the x-axis in some point of an interval [a,b], then that solution must be identically zero. Why?

I've tried to do something with the general expression for the solution.

y(x)=Ay_1(x)+By_1(x)\int \displaystyle \frac{e^{\int -P(x)dx}}{y_1^2(x)}dx \Rightarrow y'(x)=Ay_1'(x)+By_1'(x)\int \displaystyle \frac{e^{\int -P(x)dx}}{y_1^2(x)}dx +By_1(x)\displaystyle \frac{e^{\int -P(x)dx}}{y_1^2(x)}

So I've considered y'(x)=0, because y(x) must be tangent to the x axis.
0=Ay_1'(x)+By_1'(x)\int \displaystyle \frac{e^{\int -P(x)dx}}{y_1^2(x)}dx +B \displaystyle \frac{e^{\int -P(x)dx}}{y_1(x)}\Rightarrow y_1'\left( A+B \int \displaystyle \frac{e^{\int -P(x)dx}}{y_1^2(x)}dx \right) =-B \displaystyle \frac{e^{\int -P(x)dx}}{y_1(x)}

So, I don't know what to do next. I think this is not the right way.Damn, I think I got it.
y(x_0)=0, it touches the x axis
y'(x_0)=0

Thats all?

 

[LCKurtz]

Damn, I think I got it.
y(x_0)=0, it touches the x axis
y'(x_0)=0

Thats all?

Yes. By the uniqueness theorem for the initial value problem. Both y identically 0 and your supposed solution satisfy that IVP so they are the same.