two solutions in $2^{nd}$ order linear differential equations Could you please explain why we need two solutions $y_1$ and $y_2$ (fundamental set of solutions) for determine the general solution $y=cy_1+c_2y_2$ for a $2^{nd}$ order linear differential equation (probably it is the same for nonlinear case)
 A: In general, the solution of a differential equation
$$
y`=f(x,y,y',\dots,y^{(n)})
$$
of order $n$ depends on $n$ constants. One way of thinking about this is that you have to integrate $n$ times, and each integration introduces a constant. If you know the theorem on existence and uniqueness, you can see it because, under quite general conditions on $f$ and for a fixed $x_0$, for each choice of constants $\{C_1,\dots,C_n\}$ there is a unique solution such that
$$
y(x_0)=C_1,\dots, y^{(n-1)}(x_0)=C_n.
$$
For linear equations of second order
$$
y''+a(x)\,y'+b(x)\,y=0
$$
with $a$ and $b$ continuous on a neighborhood of a point $x_0$ we have the following chain of arguments:


*

*The set of solutions is a vector space.

*There are solutions $y_1$ and $y_2$ such that $y_1(x_0)=1$, $y_1'(x_0)=0$, $y_2(x_0)=0$ and $y_2'(x_0)=1$.

*$y_1$ and $y_2$ are linearly independent.

*Any solution is a linear combination of $y_1$ and $y_2$. In fact, if
$y$ is a solution then $$ y(x)=y(x_0)\,y_1(x)+y'(x_0)\,y_2(x). $$

