# Differential Equations of first order and higher degree

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.

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.
• @SaiTeja $c$ is an arbitrary constant! So $(x+y+c)(x-y+c)$ = 0 is the general solution. This equation is satisfied for both $c = c_1$ or $c = c_2$. For $c = c_1$, $x+y+c = 0$. So our general eqn. is satisfied. For $c = c_2$, $x-y+c = 0$. So again our general eqn is satisfied. We are not proposing that all $c_i$ be equal. Oct 8, 2017 at 18:36