two solutions in $2^{nd}$ order linear differential equations

Question

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)

Answer 1

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:

1. The set of solutions is a vector space.
2. 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 $y_2'(x_0)=1$.
3. $y_1$ and $y_2$ are linearly independent.
4. Any solution is a linear combination of $y_1$ and $y_2$. In fact, if $y$ is a solution then $$ y(x)=y(x_0)\,y_1(x)+y'(x_0)\,y_2(x). $$