Show that the derivative of any solution of equation $ y'' + py'+qy=0$ is also a solution.

Question

Solving $ y'' + py'+qy=0$ (where $p$ and $q$ are constants) :

Since $e^{mx}$ has the property that its derivatives are all constant multiples of the function itself, We consider $y=e^{mx}$ and substitute in the equation $ y'' + py'+qy=0$. This gives us $ ( m^2+pm+q)e^{mx}=0 $ and we solve for the roots of equation $m^2+pm+q=0$. After finding the nature of roots there can be three type of solutions possible and their formulas are derived. But the question states:

Without using the formulas obtained in this section (the solutions obtained after solving roots and substituting them), show that the derivative of any solution of equation $ y'' + py'+qy=0$ is also a solution.

Approach : Let the solution by $y$, if $y'$ is also a solution then it must satisfy $ y''' + py''+qy'=0$. Now if I solve this then it will be the same case as above but will have $ m( m^2+pm+q)e^{mx}=0 $ instead. Where do I go from here?

Answer 1

The ODE is autonomous and homogeneous. Thus if $y(x)$ is a solution, so also are the shifted functions $y(x+a)$ and the linear combinations $\frac{y(x+a)-y(x)}a$ for $a>0$ including the limit for $a\to0$ which is $y'(x)$.

Answer 2

This is trivial. But you have to get things straight. If $y$ is a solution to the DE you need to show that $y'$ is a solution - a proof of this that starts out saying "if $y'$ is a solution..." is no good at all, since it assumes what you're trying to prove.

Suppose $y$ is a solution to the DE. Just to give it a name, say $z=y'$. Now you can simply calculate $$z''+pz'+qz=\dots;$$hence $z''+pz'+qz=0$, qed.