Observe that the equation of (1) is from the algebraic point of view a secondorder equation in the variable Y={V}^{\mathrm{\prime}\mathrm{\prime}}. So, solving it algebraically in order of {V}^{\u2033}, we obtain the equivalent form
{V}^{\u2033}=\frac{p{x}^{2}\pm \sqrt{{p}^{2}{x}^{4}4q{x}^{3}(x{V}^{\prime}V)}}{2{x}^{3}},
which leads us to consider the equation
{V}^{\u2033}+{H}_{\pm}(x,V,{V}^{\prime})=0,
where
{H}_{\pm}(x,V,{V}^{\prime})=\frac{p{x}^{2}\mp \sqrt{{p}^{2}{x}^{4}4q{x}^{3}(x{V}^{\prime}V)}}{2{x}^{3}}.
This fact suggests the study of the auxiliary problem
\{\begin{array}{c}{V}^{\u2033}+H(x,V,{V}^{\prime})=0,\hfill \\ V(c)={V}_{c},\phantom{\rule{2em}{0ex}}V(d)={V}_{d},\hfill \end{array}
(2)
where
H(x,V,{V}^{\prime})=\frac{p{x}^{2}\sqrt{{p}^{2}{x}^{4}+4q{x}^{3}x{V}^{\prime}V}}{2{x}^{3}}.
The main argument to prove Theorem 2 relies on the method of upper and lower solutions.
We recall that \alpha \in {C}^{2} is a lower solution of (2) if
\{\begin{array}{c}{\alpha}^{\mathrm{\prime}\mathrm{\prime}}+H(x,\alpha ,{\alpha}^{\mathrm{\prime}})\ge 0,\hfill \\ \alpha (c)\le {V}_{c},\phantom{\rule{2em}{0ex}}\alpha (d)\le {V}_{d}.\hfill \end{array}
(3)
Similarly, an upper solution \beta \in {C}^{2} of (2) is defined by reversing the inequalities in (3). A solution of (2) is a function u which is simultaneously a lower and an upper solution. A function f is said to satisfy the Nagumo condition on some given subset E\subset I\times {\mathbb{R}}^{2} if there exists a positive continuous function \phi \in C({\mathbb{R}}_{0}^{+},[\epsilon ,+\mathrm{\infty}[), \epsilon >0, such that
f(x,y,z)\le \phi (z),\phantom{\rule{1em}{0ex}}\mathrm{\forall}(x,y,z)\in E,
and
{\int}_{0}^{+\mathrm{\infty}}\frac{s}{\phi (s)}\phantom{\rule{0.2em}{0ex}}ds=+\mathrm{\infty}.
(4)
Lemma 1
Suppose that
\frac{{V}_{d}}{d}\le \frac{{V}_{c}}{c}.
(5)
Then:

1.
The function {\alpha}_{1}(x)=\frac{{V}_{d}}{d}x is a lower solution of the problem (2).

2.
If k=\sqrt{\frac{q}{{c}^{3}}} is small enough, then the function
{\alpha}_{2}(x)=\frac{{V}_{d}{V}_{c}}{{d}^{2}{c}^{2}}{x}^{2}+\frac{{d}^{2}{V}_{c}{c}^{2}{V}_{d}}{{d}^{2}{c}^{2}}
is a lower solution of the problem (2).

3.
The function
\beta (x)=\frac{{V}_{d}{V}_{c}}{dc}x+\frac{d{V}_{c}c{V}_{d}}{dc}
is an upper solution of the problem (2).
Proof 1. If we plug {\alpha}_{1}(x)=\frac{{V}_{d}}{d}x in the first member of the equation in (2), we have that
{\left(\frac{{V}_{d}}{d}x\right)}^{\u2033}+H(x,\frac{{V}_{d}}{d}x,{\left(\frac{{V}_{d}}{d}x\right)}^{\prime})=0+\frac{p{x}^{2}\sqrt{{p}^{2}{x}^{4}4q{x}^{3}(\frac{{V}_{d}}{d}x\frac{{V}_{d}}{d}x)}}{2{x}^{3}}=0,
and by (5)
{\alpha}_{1}(c)=\frac{{V}_{d}}{d}c\le {V}_{c},\phantom{\rule{2em}{0ex}}{\alpha}_{1}(d)=\frac{{V}_{d}}{d}d={V}_{d}.

2.
Consider now the function
{\alpha}_{2}(x)=\frac{{V}_{d}{V}_{c}}{{d}^{2}{c}^{2}}{x}^{2}+\frac{{d}^{2}{V}_{c}{c}^{2}{V}_{d}}{{d}^{2}{c}^{2}}.
Observe first that
{\alpha}_{2}(c)=\frac{{V}_{d}{V}_{c}}{{d}^{2}{c}^{2}}{c}^{2}+\frac{{d}^{2}{V}_{c}{c}^{2}{V}_{d}}{{d}^{2}{c}^{2}}={V}_{c}
and
{\alpha}_{2}(d)=\frac{{V}_{d}{V}_{c}}{{d}^{2}{c}^{2}}{d}^{2}+\frac{{d}^{2}{V}_{c}{c}^{2}{V}_{d}}{{d}^{2}{c}^{2}}={V}_{d}.
On the other hand, we have that
\begin{array}{rcl}H(x,V,{V}^{\mathrm{\prime}})& =& \frac{p{x}^{2}\sqrt{{p}^{2}{x}^{4}+4q{x}^{3}x{V}^{\prime}V}}{2{x}^{3}}\\ \ge & \frac{p{x}^{2}\sqrt{{p}^{2}{x}^{4}}\sqrt{4q{x}^{3}x{V}^{\prime}V}}{2{x}^{3}}\\ =& \sqrt{\frac{q}{{x}^{3}}}\sqrt{x{V}^{\prime}V}\\ \ge & \sqrt{\frac{q}{{c}^{3}}}\sqrt{x{V}^{\prime}V}\\ =& k\sqrt{x{V}^{\prime}V},\end{array}
where k=\sqrt{\frac{q}{{c}^{3}}} .
So, if we plug the function {\alpha}_{2}(x) in the first member of the equation in (2), we obtain
\begin{array}{r}{(\frac{{V}_{d}{V}_{c}}{{d}^{2}{c}^{2}}{x}^{2}+\frac{{d}^{2}{V}_{c}{c}^{2}{V}_{d}}{{d}^{2}{c}^{2}})}^{\u2033}\\ \phantom{\rule{2em}{0ex}}+H(x,\frac{{V}_{d}{V}_{c}}{{d}^{2}{c}^{2}}{x}^{2}+\frac{{d}^{2}{V}_{c}{c}^{2}{V}_{d}}{{d}^{2}{c}^{2}},{(\frac{{V}_{d}{V}_{c}}{{d}^{2}{c}^{2}}{x}^{2}+\frac{{d}^{2}{V}_{c}{c}^{2}{V}_{d}}{{d}^{2}{c}^{2}})}^{\prime})\\ \phantom{\rule{1em}{0ex}}\ge 2\frac{{V}_{d}{V}_{c}}{{d}^{2}{c}^{2}}k\sqrt{2{x}^{2}\frac{{V}_{d}{V}_{c}}{{d}^{2}{c}^{2}}\frac{{V}_{d}{V}_{c}}{{d}^{2}{c}^{2}}{x}^{2}\frac{{d}^{2}{V}_{c}{c}^{2}{V}_{d}}{{d}^{2}{c}^{2}}}\\ \phantom{\rule{1em}{0ex}}=2\frac{{V}_{d}{V}_{c}}{{d}^{2}{c}^{2}}k\sqrt{{x}^{2}\frac{{V}_{d}{V}_{c}}{{d}^{2}{c}^{2}}\frac{{d}^{2}{V}_{c}{c}^{2}{V}_{d}}{{d}^{2}{c}^{2}}},\end{array}
which, for k small enough, is nonnegative.

3.
Consider now the function
\beta (x)=\frac{{V}_{d}{V}_{c}}{dc}x+\frac{d{V}_{c}c{V}_{d}}{dc}.
Then, in an analogous way, we plug \beta (x) in the first member of the equation in (2). We observe that
\begin{array}{r}{(\frac{{V}_{d}{V}_{c}}{dc}x+\frac{d{V}_{c}c{V}_{d}}{dc})}^{\u2033}\\ \phantom{\rule{2em}{0ex}}+H(x,\frac{{V}_{d}{V}_{c}}{dc}x+\frac{d{V}_{c}c{V}_{d}}{dc},{(\frac{{V}_{d}{V}_{c}}{dc}x+\frac{d{V}_{c}c{V}_{d}}{dc})}^{\prime})\\ \phantom{\rule{1em}{0ex}}=0+\frac{p{x}^{2}\sqrt{{p}^{2}{x}^{4}+4q{x}^{3}\frac{{V}_{d}{V}_{c}}{dc}x\frac{{V}_{d}{V}_{c}}{dc}x+\frac{d{V}_{c}c{V}_{d}}{dc}}}{2{x}^{3}}<0\end{array}
and
\beta (c)={V}_{c},\phantom{\rule{2em}{0ex}}\beta (d)={V}_{d}.
So, the three assertions of the lemma are proved. □
Remark 1 The value {K}^{\ast}=\frac{2({V}_{d}{V}_{c})}{({d}^{2}{c}^{2})\sqrt{{V}_{c}+{V}_{d}}} is a suitable upper bound for the possible values that k=\sqrt{\frac{q}{{c}^{3}}} can take in the above assertion 2. In fact, for x\in [c,d] and since \frac{{V}_{d}}{d}\le \frac{{V}_{c}}{c}, easy computations show that
{x}^{2}({V}_{d}{V}_{c})({d}^{2}{V}_{c}{c}^{2}{V}_{d})\le ({d}^{2}{c}^{2})({V}_{d}+{V}_{c}).
Therefore, if k\le {K}^{\ast}, we have
\begin{array}{rcl}k& \le & \frac{2({V}_{d}{V}_{c})}{({d}^{2}{c}^{2})\sqrt{{V}_{c}+{V}_{d}}}\\ \le & \frac{2\frac{{V}_{d}{V}_{c}}{{d}^{2}{c}^{2}}}{\sqrt{{x}^{2}\frac{{V}_{d}{V}_{c}}{{d}^{2}{c}^{2}}\frac{{d}^{2}{V}_{c}{c}^{2}{V}_{d}}{{d}^{2}{c}^{2}}}}\phantom{\rule{1em}{0ex}}\text{for all}x\in [c,d],\end{array}
which is used in the last step of the proof of assertion 2.
Now we state an existence and localisation result for the auxiliary problem (2).
Theorem 1 Suppose that \frac{{V}_{d}}{d}\le \frac{{V}_{c}}{c}. Then:

1.
The problem (2) has a solution V such that
\frac{{V}_{d}}{d}x\le V(x)\le \frac{{V}_{d}{V}_{c}}{dc}x+\frac{d{V}_{c}c{V}_{d}}{dc}.

2.
If k=\sqrt{\frac{q}{{c}^{3}}} is small enough, the problem (2) has a solution V such that
\frac{{V}_{d}{V}_{c}}{{d}^{2}{c}^{2}}{x}^{2}+\frac{{d}^{2}{V}_{c}{c}^{2}{V}_{d}}{{d}^{2}{c}^{2}}\le V(x)\le \frac{{V}_{d}{V}_{c}}{dc}x+\frac{d{V}_{c}c{V}_{d}}{dc}.

3.
If \frac{{V}_{d}}{d}=\frac{{V}_{c}}{c}, the function V(x)=\frac{{V}_{d}}{d}x is a solution of the problem (2).
Proof By the previous lemma, we know already that there are lower and upper solutions for the problem (2). It is also clear that they are well ordered, that is,
{\alpha}_{1}\le \beta \phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\alpha}_{2}\le \beta .
So, if the function H satisfies the Nagumo condition, the thesis will follow by [1] or [2].
In order to prove assertions 1 and 2, we consider the sets, respectively,
{E}_{1}=\{(x,y,z)\in [c,d]\times {\mathbb{R}}^{2}:\frac{{V}_{d}}{d}x\le y\le \frac{{V}_{d}{V}_{c}}{dc}x+\frac{d{V}_{c}c{V}_{d}}{dc}\}
and
{E}_{2}=\{(x,y,z)\in [c,d]\times {\mathbb{R}}^{2}:\frac{{V}_{d}{V}_{c}}{{d}^{2}{c}^{2}}{x}^{2}+\frac{{d}^{2}{V}_{c}{c}^{2}{V}_{d}}{{d}^{2}{c}^{2}}\le y\le \frac{{V}_{d}{V}_{c}}{dc}x+\frac{d{V}_{c}c{V}_{d}}{dc}\}.
Function H satisfies the Nagumo condition in {E}_{1} and in {E}_{2}. In fact, we have that
\begin{array}{rl}H(x,y,z)& =\left\frac{p{x}^{2}\sqrt{{p}^{2}{x}^{4}+4q{x}^{3}xzy}}{2{x}^{3}}\right\\ \le \frac{p{x}^{2}+\sqrt{4q{x}^{3}xzy}}{2{x}^{3}}\\ \le \frac{p}{2x}+\sqrt{\frac{q}{{x}^{3}}}\sqrt{\frac{{V}_{d}{V}_{c}}{dc}x+\frac{d{V}_{c}c{V}_{d}}{dc}}+\frac{\sqrt{q}}{x}\sqrt{z}\\ \le \frac{p}{2c}+\sqrt{\frac{q}{{c}^{3}}}\sqrt{\frac{{V}_{d}{V}_{c}}{dc}d+\frac{d{V}_{c}c{V}_{d}}{dc}}+\frac{\sqrt{q}}{c}\sqrt{z}.\end{array}
Thus, for some positive constants {k}_{1} and {k}_{2}, we have that
H(x,y,z)\le {k}_{1}+{k}_{2}\sqrt{z},\phantom{\rule{1em}{0ex}}\mathrm{\forall}(x,y,z)\in {E}_{1}\cup {E}_{2}
and
{\int}_{0}^{+\mathrm{\infty}}\frac{s}{{k}_{1}+{k}_{2}\sqrt{s}}\phantom{\rule{0.2em}{0ex}}ds=+\mathrm{\infty}.
So, by what was said above, the first two assertions of the thesis hold.
The third assertion follows directly from obvious computations. □
Proposition 1 Consider the problem (2) and the solution V given by Theorem 1. Then V is convex and satisfies V(x)\ge x{V}^{\mathrm{\prime}}(x) in [c,d].
Proof The convexity of the solution V follows easily by the equation of the problem (2)
{V}^{\u2033}=H(x,V,{V}^{\prime})=\frac{p{x}^{2}+\sqrt{p{x}^{4}+4q{x}^{3}x{V}^{\prime}V}}{2{x}^{3}}\ge 0.
Noting that
{(x{V}^{\prime}(x)V(x))}^{\prime}=x{V}^{\u2033}(x)\ge 0
then
x{V}^{\prime}(x)V(x)\le d{V}^{\prime}(d){V}_{d}.
But, using Theorem 1, assertion 1, we have that {V}^{\prime}(d)\le \frac{{V}_{d}}{d}. In fact, since \frac{{V}_{d}}{d}x\le V(x), it follows that \frac{{V}_{d}}{d}\ge \frac{V(x){V}_{d}}{xd}, for c\le x<d. Letting x\to d, we obtain that \frac{{V}_{d}}{d}\ge {V}^{\mathrm{\prime}}(d).
Then we have
x{V}^{\prime}(x)V(x)\le 0.
□