I have a very simple linear first order ODE. 


$$v(x) = c x + A - B x(1-x) v'(x)$$ 


$c, A ,B \in(0,1)$. The domain is $(\underline{x}, 1)$. where $\underline x > 0$. 


I am guessing that for any initial condition $v(\underline x)$ there will be a unique solution. In particular, suppose I give an initial condition $v(\underline x)$ such that $0 < v(\underline x) < c \underline x + A$. I would like to argue that the solution must be convex. What are the sort of methods that I can use when an explicit closed form solution isn't available? 


Thanks 



Convexity of a solution of a first order linear ODE
If I did not make any mistake, $v(x)$ need not be convex.




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We find that
$$B v'(x) = \frac{c x + A - v(x)}{x (1 - x)} ,$$
and therefore
$$B v''(x) = \frac{(c - v'(x)) x (1 - x) - (c x + A - v(x)) (1 - 2 x)}{x^2 (1 - x)^2} .$$
Plugging in the expression for $v(x)$, we get
$$B v''(x) = \frac{(c - v'(x)) x (1 - x) + B x (1 - x) v'(x) (1 - 2 x)}{x^2 (1 - x)^2} .$$
which leads to
$$B v''(x) = \frac{c - (1 + B (1 - 2 x)) v'(x)}{x (1 - x)} .$$
This is positive a long as $(1 + B(1 - 2 x)) v'(x) < c$. This need not be satisfied at the initial condition if $\underline{x}$ is small enough: then $1 + B(1 - 2 \underline{x}) > 0$ and $v'(\underline{x})$ can be arbitrarily close to
$$
 \frac{c \underline{x} + A}{B \underline{x} (1 - \underline{x})} ,
$$
which can be arbitrarily large as $\underline{x} \to 0^+$.


