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I've given the vague ODE: $$x''(t) + a_1\,x'(t)+a_0\,x(t) =0$$

gathered with the premade solution $$\displaystyle{x(t) = b_1\,e^{\gamma_1\,t}+b_2\,e^{\gamma_2\,t}}$$

The task is to determine how $\gamma_1, \gamma_2$ depend on $a_0, a_1$ by comparison of coefficients

My first try was to differentiate the solution:

$$\left(b_1\,\gamma_1^2\,e^{\gamma_1\,t}+b_2\,\gamma_2^2\,e^{\gamma_2\,t}\right)+a_1\,\left(b_1\,\gamma_1\,e^{\gamma_1\,t}+b_2\,\gamma_2\,e^{\gamma_2\,t}\right)+a_0\,\left(b_1\,\,e^{\gamma_1\,t}+b_2\,\,e^{\gamma_2\,t}\right) = 0$$

But since I didn't really glimpsed a way to simplify that expression. I started vice versa and solving the ODE: $$\begin{align}&x''(t) + a_1\,x'(t)+a_0\,x(t) =0 \quad\Rightarrow \lambda^2+a_1\,\lambda+a_0 \\[12pt] &\text{hence:}\quad \lambda_{1,2} = \frac{a_1}{2}\pm\sqrt{\frac{a_1^2}{4}-a_0} \quad = \gamma_{1,2}\end{align}$$

What brings straight the dependency. Now to my question: Is there a way to arrive at the result by the method above?

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  • $\begingroup$ Now I saw it: if you group all terms with $b_1$ on one side: $b_1\,\gamma_1^2+a_1\,b_1\,\gamma+a_0\,b_1 = 0$ you get back the result. (Likewise $b_2$) $\endgroup$
    – Leon
    Commented May 21, 2021 at 17:28

1 Answer 1

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Collect terms in $e^{\gamma_1 t}$ and $e^{\gamma_2 t}$ in your equation $$\left(b_1\,\gamma_1^2\,e^{\gamma_1\,t}+b_2\,\gamma_2^2\,e^{\gamma_2\,t}\right)+a_1\,\left(b_1\,\gamma_1\,e^{\gamma_1\,t}+b_2\,\gamma_2\,e^{\gamma_2\,t}\right)+a_0\,\left(b_1\,\,e^{\gamma_1\,t}+b_2\,\,e^{\gamma_2\,t}\right) = 0$$ In order for this to be true for all $t$ (if $\gamma_1 \ne \gamma_2$), the coefficients of $e^{\gamma_1 t}$ and $e^{\gamma_2 t}$ must both be $0$.

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  • $\begingroup$ thank you, but it seems unlikely to find $\gamma_{1,2} = \frac{a_1}{2}\pm\sqrt{\frac{a_1^2}{4}-a_0}$ just by doing so $\endgroup$
    – Leon
    Commented May 21, 2021 at 17:04

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