# comparison of coefficient in dummy-ODE

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?

• 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 $b_2$)
– Leon
Commented May 21, 2021 at 17:28

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 $$0$$.
• 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