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Would this be considered sufficient/correct proof for a second solution of 2nd order homogeneous linear differential equations? I skipped some steps and explanations here just for brevity:

$$(1) \ \ \ ay''+by'+cy=0$$ The characteristic equation of $(1)$ is: $$(2) \ \ \ ar^{2}+br+c=0$$ Assuming $b^{2}-4ac=0$ then $(2)$ has roots $r_1=r_2=r=\frac{-b}{2a}$, and so $y_1=e^{rt}$ is a solution,

$$\implies\ a\frac{d^{2}}{dt^{2}}\left[ e^{rt} \right]+b\frac{d}{dt}\left[ e^{rt} \right]+ce^{rt}=0$$ Multiplying both sides by $\frac{d}{dr}$ we get: $$a\frac{d^{2}}{dt^{2}}\left[ te^{rt} \right]+b\frac{d}{dt}\left[ te^{rt} \right]+cte^{rt}=0$$ $\implies \ y_{2}=te^{rt}$ is a second solution satisfying $(1)$.

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  • $\begingroup$ It seems quite incomplete. You have nowhere explicitly used the fact that $r$ is a double root of the characteristic equation. $\endgroup$
    – André Nicolas
    May 10, 2016 at 5:57
  • $\begingroup$ @AndréNicolas i edited. is it better or ? $\endgroup$
    – inspd
    May 10, 2016 at 6:14
  • $\begingroup$ Where do you get this approach from, I have never seen that ? Anyway, there is no justification for differentiating like you do, as the variables $a,b,c,r$ are not independent. $\endgroup$
    – user65203
    May 10, 2016 at 6:21
  • $\begingroup$ @YvesDaoust the approach is my misinformed self lol! but ill try explaining what i was thinking back there: $a,b,c$ are constants, I'll use the first term as an example: $$\frac{d}{dr}\left[ a\frac{d^{2}}{dt^{2}}\left( e^{rt} \right) \right]=a\frac{d}{dr}\left[ \frac{d^{2}}{dt^{2}}\left( e^{rt} \right) \right]=a\frac{d^{2}}{dt^{2}}\left[ \frac{d}{dr}\left( e^{rt} \right) \right]=a\frac{d^{2}}{dt^{2}}\left[ te^{rt} \right]$$. I did this to every term i guess $\endgroup$
    – inspd
    May 10, 2016 at 6:34
  • $\begingroup$ You were unlucky that this worked. You are missing terms $\partial a/\partial r,\partial b/\partial r,\partial c/\partial r$. $\endgroup$
    – user65203
    May 10, 2016 at 6:49

1 Answer 1

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By the method of variation of the constant, let

$$y_2=C(t)y_1(t)$$ and plug in the equation.

$$a(C''(t)y_1(t)+2C'(t)y_1'(t)+C(t)y_1''(t))+b(C'(t)y_1(t)+C(t)y_1'(t))+cC(t)y_1(t)=0$$

gives after simplification (expressing that $y_1$ is a solution)

$$a(C''(t)y_1(t)+2C'(t)y_1'(t))+bC'(t)y_1(t)=0,$$

$$aC''(t)e^{rt}+2aC'(t)re^{rt}+bC'(t)e^{rt}=0,$$

$$aC''(t)+(2ar+b)C'(t)=0.$$

As $r$ is a double root, $2ar+b=0$, and a particular solution of $C''(t)=0$ is

$$C(t)=t,\\y_2(t)=te^{rt}.$$

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