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Consider the first order linear ODE

$F(x) + A x F'(x) + B = 0$, with initial condition $F(1) = 1$. Moreover, let $A,B \neq 0$, real, and $sgn(A) = sgn(B)$.*

*... if the latter matters.

Mathematica gives me the following solution, which is easy to verify:

$F(x) = (1+B)(x^{-1/A}) - B$.

My question is: is there any elementary way or theorem to establish that the above must be the unique global solution to the given ODE? Do I have to apply the Picard–Lindelöf theorem (if that works), or is there an easier argument?

I have not much knowledge about how to solve differential equations; the question arises from an applied economics paper I'm working on.

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You can't expect "global" uniqueness (for all real $x$). The reason is that the differential equation provides no information about $F'(0)$ and thus the portions for $x > 0$ and for $x < 0$ "don't know about each other".

Here's how to prove uniqueness on $(0, \infty)$.Suppose you have two solutions $F$ and $G$ of the same problem. Set $f(x) = F(x) - G(x)$ for $x > 0$, then $Axf'(x) + f(x) = 0$ and $f(1) = 0$. Now set $g(x) = x^{1/A}f(x)$. Then you can check that $g'(x) = 0$ for $x > 0$ and $g(1) = 0$. Therefore $g(x) = 0$ for all $x >0 $ and therefore $F(x) = G(x)$ for all $x > 0$.

Here is an example for two solutions that agree for $x > 0$ but are different for $x < 0$. Consider the two functions $$ F(x) = \begin{cases} (1+B)x^{-1/A} - B \quad (x > 0)\\ B x^{-1/A} - B \quad (x < 0) \end{cases} $$ and $$ G(x) = (1+B)x^{-1/A} - B \quad (x \ne 0) \, . $$ Set $F(0) = G(0) = -B$. Both functions satisfy the differential equation and the condition $F(1) = 1$, but they differ for $x < 0$.

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  • $\begingroup$ Many thanks for providing this elegant proof. I particularly like the trick of specifying $g(x) = x^{1/A)f(x)$ in order to exploit the properties of the product rule. Also, it's good to know that the solution is in fact unique on the domain $x > 0$, as that is all that matters for my problem. $\endgroup$ – Martin Jun 4 '13 at 15:40

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