Let $F\in C([a,b]\times\mathbb{R}^{3}).$ We consider the following implicit differential equation

By definition this equation is homogeneous if the function $F$ is homogeneous with respect to the last three arguments.

In this paper by a solution of the equation (1.1) we understand a function $y\in C^{2}[a,b]$ which satisfies (1.1). Moreover by a solution we shall understand a nontrivial solution.

The linear case of (1.1) is the following equation

For the equation (1.2) the following properties of the solution are well known (see [10], [11], [12], [7], [3], [13]).

We suppose that $p$ and $q\in C[a,b].$

The aim of this paper is to extend the above results to the solutions of (1.1). For some results in this directions see [14], [8] and [9].

We shall use the above examples to exemplify our general results.

For other examples of such equations see [4] and [12].

We consider the equation (1.1) with $F\in C([a,b]\times\mathbb{R}^{3})$ a homogeneous function with respect to the last three arguments. We have the following extremal principle.

• (a)

  Let $x_{0}\in]a,b[$ be such that, $y(x_{0})>0$ is the maximum value of $y$ on $[a,b]$. Since $y\in C^{2}[a,b]$ we have that $y(x_{0})>0,y^{\prime}(x_{0})=0,y^{\prime\prime}(x_{0})\leq 0$. From (1.1) we have

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  $$F(x_{0},y(x_{0}),0,y^{\prime\prime}(x_{0}))=0.$$

  But in the condition of our theorem the first part of this relation is not equal to zero. So, $x_{0}\in\{a,b\}.$

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• (b)

  We remark that if $y$ is a solution of (1.1) then, $-y$ is also a solution. We apply $(a)$ for $-y$.

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∎

Now let us consider the bilocal problem

We have for this problem the following result.

Follows from Theorem 5. ∎

Now we consider the following conditions on (1.1):

• (u₀)

  If $y$ is a solution of (1.1) and for some $x_{0}\in[a,b],\ y(x_{0})=0$ and $y^{\prime}(x_{0})=0,$ then $y=0$.

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• (u₁)

  If $y_{1}$ and $y_{2}$ are solutions of (1.1) and for some $x_{0}\in[a,b],\ y_{1}(x_{0})=y_{2}(x_{0})>0$ and $y_{1}^{\prime}(x_{0})=y_{2}^{\prime}(x_{0}),$ then $y_{1}=y_{2}$.

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• (u₂)

  If $y$ is a solution of (1.1) and for some $x_{0}\in[a,b],\ y^{\prime}(x_{0})=0$, $y^{\prime\prime}(x_{0})=0,$ then $y=0$.

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• (u₃)

  If $y_{1}$ and $y_{2}$ are solutions of (1.1) and for some $x_{0}\in[a,b],\ y_{1}^{\prime}(x_{0})=y_{2}^{\prime}(x_{0})>0$ and $y_{1}^{\prime\prime}(x_{0})=y_{2}^{\prime\prime}(x_{0}),$ then $y_{1}=y_{2}$.

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By standard arguments we have

In what follow we also need the following result (see [14], [10], p. 163 and [5]).

Using Lemma 3, Tonelli give in [14] the following result.

Our results are the following.

We consider $x_{1}$ and $x_{2}$ two consecutive zeros of $y^{\prime}(x)$. We have to prove that $y(x)$ has at least one zero in the interval $(x_{1},x_{2})$.

We suppose that $y(x)\neq 0,\ x\in[x_{1},x_{2}]$. Applying Tonelli’s Lemma 3 there exists $x_{0}\in(x_{1},x_{2})$ and $\lambda>0\ ($ or $<0)$ such that

So,

Using (4.1) in $F(x_{0},y(x_{0}),y^{\prime}(x_{0}),y^{\prime\prime}(x_{0}))=0$ we obtain that ($y(x_{0})\neq 0$)

∎

We consider $x_{1}$ and $x_{2}$ two consecutive zeros of $y_{1}^{\prime}(x)$. We have to prove that $y_{2}^{\prime}(x)$ has at least one zero in $[x_{1},x_{2}]$.

We suppose that $y_{2}^{\prime}(x)\neq 0,\ x\in[x_{1},x_{2}]$. Applying Tonelli’s Lemma 3 there exists $x_{0}\in(x_{1},x_{2})$ and $\lambda>0\ ($ or $<0)$ such that

Taking into account (u₂) we have that $y_{2}(x)=\lambda y_{1}(x)$ and so we have reached a contradiction. ∎

The problem is to study the above problem for the equation (1.1).