# Why is the sum of two independent solutions of homogenous 2nd order ODEs a solution?

I am reading on the solutions of second order homogenous ODEs (linear), and came across this: Now, my questions is split into two parts:

1) I know that the independent solutions are equations, however are they solutions to same differential equation (I mean to say, "are they both solutions derived from the second integration"), or for example, is $y_1$ the solution of the first integration and $y_2$ the solution of the second?

2)Now, if these independent solutions are indeed the solutions to the same differential equations, or the above example is true, or whatever else the independent solutions resemble, why is it that their sum is also a solution?

1. The two independent solutions are functions, not equations. The functions can (sometimes) be defined by equations. Both functions are solutions of the same equation. This is very much analogous to having two solutions $x$ to $x^2-1=0$.
2. If $y_1$ and $y_2$ are functions $\mathbb R\to\mathbb R$, so is $y_1+y_2$. This function is defined via $(y_1+y_2)(t)=y_1(t)+y_2(t)$. Checking that $y_1+y_2$ also solves the differential equation is a calculation that I urge you to do yourself. (The underlying reason is that the equation is linear, but you need not worry about that here. You can just calculate.)
• @ReinhildVanRosenú, you don't need to have explicit formulas for the functions $y_1$ and $y_2$. You can still calculate. The derivative of $y_1+y_2$ is $y_1'+y_2'$ and similarly for the second derivatives. If we denote $y=y_1+y_2$, you can just start calculating $ay''+by'+c$ using what you know and see if you end up with zero. – Joonas Ilmavirta Jul 7 '15 at 20:26
• I managed to get $a(y_1''+y_2'') + b(y_1'+y_2')+c$with substitution, which is not equal to zero. Can you guide me as to which property I have to use, I have just began properly studying ODEs today. – Reinhild Van Rosenú Jul 7 '15 at 20:35
• @ReinhildVanRosenú, you should get $a(y_1''+y_2'')+b(y_1'+y_2')+c(y_1+y_2)$, which in turn equals $[ay_1''+by_1'+cy_1]+[ay_2''+by_2'+cy_2]$. – Joonas Ilmavirta Jul 7 '15 at 20:37