Now the pieces fit together: derivatives become multiplication by $s$, the equation becomes algebra, and inverse transforms bring the answer back. The initial conditions enter at the start, so there is no general solution to fit afterwards.
Idea clave
Transform both sides (using $\mathcal{L}\{y'\} = sY - y(0)$, $\mathcal{L}\{y''\}$ $= s^{2}Y - sy(0) - y'(0)$), solve the algebraic equation for $Y(s)$, split it into partial fractions, and invert.
Por qué funciona
For $ay'' + by' + cy = g$, transforming gives $(as^{2} + bs + c)Y$ $= G(s)$ $+ (\text{terms from the initial values})$. The polynomial multiplying $Y$ is the characteristic polynomial, so its roots, the poles of $Y$, are exactly the exponents in the homogeneous solution. The poles of $G$ add the particular solution. Partial fractions sorts everything into pieces you can invert.
Ejemplo resuelto: First order
Solve $y' + 2y = 6$, $y(0) = 1$.
- Transform: $sY - 1 + 2Y = \frac{6}{s}$, so $Y = \frac{1}{s + 2} + \frac{6}{s(s + 2)}$.
- Partial fractions: $\frac{6}{s(s + 2)} = \frac{3}{s} - \frac{3}{s + 2}$.
- $Y = \frac{3}{s} - \frac{2}{s + 2}$.
Respuesta: $y = 3 - 2e^{-2t}$
Ejemplo resuelto: Second order
Solve $y'' - y' - 2y = 0$, $y(0) = 1$, $y'(0) = 5$.
- $(s^{2}Y - s - 5) - (sY - 1) - 2Y$ $= 0$, so $(s^{2} - s - 2)Y = s + 4$.
- $Y = \frac{s + 4}{(s - 2)(s + 1)}$ $= \frac{2}{s - 2} - \frac{1}{s + 1}$ (cover-up: $\frac{6}{3} = 2$ and $\frac{3}{-3} = -1$).
- Invert.
Respuesta: $y = 2e^{2t} - e^{-t}$, the same answer as the characteristic-equation method
Ejemplo resuelto: A constant force
Solve $y'' + 4y = 8$, $y(0) = 0$, $y'(0) = 0$.
- $(s^{2} + 4)Y = \frac{8}{s}$, so $Y = \frac{8}{s(s^{2} + 4)}$.
- Partial fractions: $\frac{8}{s(s^{2} + 4)} = \frac{2}{s} - \frac{2s}{s^{2} + 4}$.
Respuesta: $y = 2 - 2\cos 2t$
Error común
Inverting a product piece by piece, as if $\mathcal{L}^{-1}\left\{\frac{6}{s(s + 2)}\right\}$ were $6\cdot 1\cdot e^{-2t}$. The inverse of a product is not the product of inverses; split into partial fractions first (or use convolution, coming soon). Also include the initial-value terms when you transform.
When is this better than the characteristic-equation method? For smooth forcing like $e^{t}$ or $\cos t$ the two methods take about the same effort. The transform method shines when the forcing switches on and off or acts in an instant (the next two topics), because it needs no separate solution on each piece. It assumes the initial values are given at $t = 0$; for conditions at $t = 5$, shift time first with $\tau = t - 5$.