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Suppose we know all functions $\phi$ that are solutions of the differential equation

$$G(x)=F_1(x,y(x),y'(x),\dots,y^{(n)}(x))=0$$

for some arbitrary function $F_1$. Clearly those same $\phi$ also satisfy the differential equation

$$G'(x)=F_2(x,y(x),y'(x),\dots,y^{(n+1)}(x))=0$$

obtained by "differentiating the differential equation". Intuitively, I assume new solutions are generated, but I'm not certain if that is always the case and if there is a systematic way to relate those new solutions to the ones we already have for the original differential equation and to the functional form of $F_1$ and $F_2$. For example, the solutions to the differential equation

$$y'=0$$

are of the form $y(x)=C$ for some constant $C$. "Differentiating" the differential equation, we get

$$y''=0$$

which has solutions of the form $y(x)=C_1x+C_2$ for some constants $C_1$ and $C_2$. The set of solutions of the second differential equation contains the solutions of the first one, as expected, but also contains new solutions, which in this case are antiderivatives of the original solutions. It doesn't seem that things work out always as they do for this simple case, though.

Is there a systematic way of knowing what will happen to the solution set of a differential equation if you "differentiate" it?

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    $\begingroup$ @SammyBlack I made a point of defining two different functions because after differentiation of the equation we would usually get an equation of a different form. For example: $y^2=0$ leading to $2yy'=0$, which are of different forms. As to your first question: this same example shows why I still included $y$ as an argument in $F_2$. It might still appear in the equation after it has been differentiated, in addition to a derivative of higher order. $\endgroup$
    – jvf
    Aug 5 at 0:45
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    $\begingroup$ Oh, I see what you're asking. Starting with $F_1(x, y(x), y'(x), \dots, y^n(x)) = 0$, you differentiate with respect to $x$, then collect terms and the new equation is of the form $F_2(x, y(x), y'(x), \dots, y^{(n+1)}(x)) = 0$, an ODE of degree $1$ higher. $\endgroup$ Aug 5 at 0:45
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    $\begingroup$ @T_M check out my reply to SammyBlack's comment; I believe it clarifies things. $\endgroup$
    – jvf
    Aug 5 at 0:48
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    $\begingroup$ As a slight aside, in this case, the solutions will be the solutions of when $F_1(x,y(x),y'(x),\dots,y^{(n)}(x))$ is a constant. For example, the solutions of $F_2=y''=0$ are the same as the solutions to $F_1=y'=c$; both have the solution $y=cx+C_1$. This can also be seen by integrating both sides. $\endgroup$ Aug 5 at 2:01
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    $\begingroup$ @VarunVejalla I hadn't thought of that! Seems so obvious now. If I have an equation of the form $F_1=0$, then the solutions of $F_2=0$ (the "derivative" of the equation) are the solutions of $F_1=C$ for any constant $C$, on account of simple integration. $\endgroup$
    – jvf
    Aug 5 at 17:12

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