For constant coefficients, $ay'' + by' + cy = 0$, the derivatives of the unknown must cancel against the function itself. Exponentials are the functions whose derivatives are multiples of themselves, so we try $y = e^{rt}$, exactly as in checking a solution and in superposition.
Idea clave
Substituting $y = e^{rt}$ gives the characteristic equation $$ar^{2} + br + c = 0.$$ If it has two distinct real roots $r_1 \ne r_2$, the general solution is $y = c_1e^{r_1t} + c_2e^{r_2t}$.
Por qué funciona
With $y = e^{rt}$: $y' = re^{rt}$ and $y'' = r^{2}e^{rt}$, so $ay'' + by' + cy$ $= (ar^{2} + br + c)e^{rt}$. Since $e^{rt} \ne 0$, this is zero exactly when $r$ solves the characteristic equation. Two different roots give two solutions with $W = (r_2 - r_1)e^{(r_1 + r_2)t} \ne 0$, so they are independent and superposition gives every solution.
Ejemplo resuelto: A standard initial value problem
Solve $y'' + y' - 6y = 0$, $y(0) = 1$, $y'(0) = 7$.
- $r^{2} + r - 6 = (r - 2)(r + 3)$ $= 0$: $r = 2, -3$.
- $y = c_1e^{2t} + c_2e^{-3t}$, $y' = 2c_1e^{2t} - 3c_2e^{-3t}$.
- $c_1 + c_2 = 1$ and $2c_1 - 3c_2 = 7$, so $c_1 = 2$, $c_2 = -1$.
Respuesta: $y = 2e^{2t} - e^{-3t}$
Ejemplo resuelto: A leading coefficient
Solve $2y'' - 5y' + 2y = 0$, $y(0) = 0$, $y'(0) = 3$.
- $2r^{2} - 5r + 2$ $= (2r - 1)(r - 2) = 0$: $r = \frac{1}{2}, 2$.
- $y = c_1e^{t/2} + c_2e^{2t}$; $c_1 + c_2 = 0$ and $\frac{1}{2}c_1 + 2c_2 = 3$.
- Substituting $c_1 = -c_2$: $\frac{3}{2}c_2 = 3$, so $c_2 = 2$, $c_1 = -2$.
Respuesta: $y = 2e^{2t} - 2e^{t/2}$
Ejemplo resuelto: A zero root
Solve $y'' + 3y' = 0$, $y(0) = 4$, $y'(0) = -3$, and find $\lim_{t\to\infty} y$.
- $r^{2} + 3r = r(r + 3) = 0$: $r = 0, -3$. The root 0 gives the constant solution $e^{0t} = 1$.
- $y = c_1 + c_2e^{-3t}$; $c_1 + c_2 = 4$ and $-3c_2 = -3$, so $c_2 = 1$, $c_1 = 3$.
Respuesta: $y = 3 + e^{-3t} \to 3$
Error común
Reading the roots with the wrong sign: $(r - 2)(r + 3) = 0$ means $r = 2$ and $r = -3$, so the solutions are $e^{2t}$ and $e^{-3t}$. A quick check: substitute one exponential back; $4 + 2 - 6 = 0$ confirms $r = 2$. Also keep both constants: $e^{2t} + e^{-3t}$ alone cannot match two arbitrary initial values.