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Suppose we have a homogenous linear first-order ordinary differential equation:

$$ y'(x) + p(x)y(x) = 0 \, \forall x \in J \qquad(1) $$

where $J$ is an interval and $p$ is continuous on $J$. It seems to me quite intuitive that if $y(x)$ is a solution of this ODE and $\exists x_0 \in J: y(x_0)=0$, then $y(x)$ must be identically zero on $J$, i.e.: $\forall x \in J: f(x) = 0$. However, I can't quite figure out a proof of this. Does anybody know how this would go?

To be concise: by a solution of $(1)$, I mean a function $y: J\subseteq A \to \mathbb{R}$ so that $y$ is differentiable on $J$ and satisfies $(1)$.

To be proven

Let $y(x)$ be a solution of $(1)$ and $\exists x_0 \in J: y(x_0) = 0$. Then $\forall x\in J: y(x) = 0$.

Answer (suggested by Did)

Consider the function

$$ z(x) := y(x) e^{\int_{x_0}^x p(t)dt}, $$

which is defined for all $x \in J$. Then, since $y'(x) = -p(x)y(x)$:

$$ z'(x) = y'(x) e^{\int_{x_0}^x p(t)dt} + y(x)p(x)e^{\int_{x_0}^x p(t)dt}\\ = -p(x)z(x) + p(x)z(x) = 0. $$

Therefore, we have that $\forall x \in J: z(x) = z(x_0) = y(x_0)e^{\int_{x_0}^{x_0} p(t)dt} = 0$. Now, we can say: $\forall x \in J: y(x) = z(x)e^{-\int_{x_0}^x p(t)dt} = 0.$

This concludes the proof.

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up vote 1 down vote accepted

Consider the function $z:J\to\mathbb R$ defined as $$ z(x)=y(x)\exp\left(\int_{x_0}^xp(t)\mathrm dt\right). $$ The hypotheses are that $z(x_0)=0$ and, for every $x$ in $J$, $z'(x)=$ $____$, hence the function $z$ is $______$, which proves that $________$ for every $x$ in $J$.

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Thank you very much! That helped brilliantly. For reference, I will add the complete answer above. – Eric Spreen Jan 20 '13 at 15:40

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