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?

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    $\begingroup$ Differentiate the equation throughout. What do you get? $\endgroup$ – Dylan Jul 14 '18 at 15:23
  • $\begingroup$ Thanks I got it. I was thinking it would take a lot of calculations to solve it but in the meantime I forgot my basics. $\endgroup$ – Seth Rollins Jul 14 '18 at 15:49

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 $y'(x)$.

  • $\begingroup$ Yes, that's true, but the fact that the equation is preserved under limit operation requires some explanation, I think. $\endgroup$ – user539887 Jul 14 '18 at 19:52
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    $\begingroup$ The nice thing is, this argument does not operate on the equation level, just on the properties of the solution family. One can say it considers elements of the tangent set of the solution family, and as that family is a linear subspace, the tangents are part of the space. $\endgroup$ – Lutz Lehmann Jul 14 '18 at 20:39
  • $\begingroup$ A very nice explanation. $\endgroup$ – user539887 Jul 15 '18 at 20:24

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.

  • $\begingroup$ Sorry, I forgot to mention $p$ and $q$ are constants. I have made the edit. Can you please elaborate what you did in last three lines? $\endgroup$ – Seth Rollins Jul 14 '18 at 15:01
  • $\begingroup$ @SethRollins I''ve already come really close to doing your homework for you. Figure it out! If you substitute $y'=z$ then what do you get for $z''+pz'+qz$? $\endgroup$ – David C. Ullrich Jul 14 '18 at 15:11
  • $\begingroup$ @SethRollins Hint: The derivative of $0$ is $0$. $\endgroup$ – David C. Ullrich Jul 14 '18 at 15:12
  • $\begingroup$ Thanks for the help, I figured it out after reading the comment by @Dylan. It was not a homework question, I was just practising some questions from textbook. Although now I see it was really foolish to ask such a easy question. $\endgroup$ – Seth Rollins Jul 14 '18 at 15:52
  • $\begingroup$ In the answer there is a tacit assumption that one can differentiate $z$ twice: of course, it follows from the fact that all solutions are of the form polynomial times exponential (perhaps with complex coefficient). $\endgroup$ – user539887 Jul 14 '18 at 19:51

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