In my differential equations book, I have found the following:
Let $ P_0(\frac{dy}{dx})^n+P_1(\frac{dy}{dx})^{n-1}+P_2(\frac{dy}{dx})^{n-2}+……+P_{n-1}(\frac{dy}{dx})+P_n =0$ be the differential equation of first degree 1 and order n (where $P_i$ $\forall$ i $\in {0,1,2,…n}$ are functions of x and y).
Assuming that it is solvable for p, it can be represented as:
$$[p-f_1(x,y)] [p-f_2(x,y)] [p-f_3(x,y)]……..[p-f_n(x,y)] = 0$$
equating each factor to Zero, we get n differential equations of first order and first degree.
$$[p-f_1(x,y)]=0,\space [p-f_2(x,y)]=0,\space [p-f_3(x,y)]=0,\space……..[p-f_n(x,y)] = 0$$
Let the solution to these n factors be:
$$F_1(x,y,c_1)=0,\space F_2(x,y,c_2)=0,\space F_3(x,y,c_3)=0,\space…….. F_n(x,y,c_n) = 0$$
Where $c_1, c_2,c_3…..c_n$ are arbitrary constants of integration. Since all the c’s can have any one of an infinite number of values, the above solutions will remain general if we replace $c_1, c_2,c_3…..c_n$ by a single arbitrary constant c. Then the n solutions (4) can be re-written as
$$F_1(x,y,c)=0,\space F_2(x,y,c)=0,\space F_3(x,y,c)=0,\space…….. F_n(x,y,c) = 0$$ They can be combined to form the general solution as follows:
$$F_1(x,y,c)\space F_2(x,y,c)\space F_3(x,y,c)\space…….. F_n(x,y,c) = 0\space \space \space \space \space \space\space \space \space\space \space \space(1)$$
Now, my question is, whether equation (1) is the most general form of solution to the differential equation.
I think the following is the most general form of solution to the differential equation :
$$F_1(x,y,c_1)\space F_2(x,y,c_2)\space F_3(x,y,c_3)\space…….. F_n(x,y,c_n) = 0\space \space \space \space \space \space\space \space \space\space \space \space(2)$$
If (1) is the general solution, the constant of integration can be found out by only one IVP say, $y(0)=0$. So, one IVP will give the particular solution. If (2) is the general solution, one IVP might not be able to give the particular solution to the problem.
Best Answer
Indeed equation $1$ is the general solution! Notice that equation $1$ is satisfied even if any one of $F_i(x,y,c_i)$ is zero.
Suppose that you have $c = c_1$. Then equation $1$ is satisfied because $F_1 =0$. If you have $c = c_2$, then again eq. $1$ is satisfied because $F_2 = 0$. And so on.
Your equation $2$ is a special solution, as in this case, all $F_1, F_2 \dots F_n$ are individually zero.