Question

Solve y"-5y'-6y=e^(-2t) where y’(0) = 1 and y(0) = -1 using LaPlace Transforms

Answer 1

Solve $y'' - 5y' - 6y = e^{-2t}$, where $y'(0) = 1$ and $y(0) = -1$ using Laplace Transform.

**Solution:**

$$y \to Y(p)$$ $$y' \to pY(p) - y(0) = pY(p) + 1$$ $$y'' \to p^2Y(p) - py(0) - y'(0) = p^2Y(p) + p - 1$$ $$e^{-2t} \to \frac{1}{p + 2}$$ $$\begin{aligned} p^2Y(p) + p - 1 - 5(pY(p) + 1) - 6Y(p) &= \frac{1}{p + 2} \to (p^2 - 5p - 6)Y(p) \\ &= \frac{1}{p + 2} - p + 6 \end{aligned}$$ $$\begin{aligned} (p^2 - 5p - 6)Y(p) &= \frac{-p^2 + 4p + 13}{p + 2} \to Y(p) = \frac{-p^2 + 4p + 13}{(p + 2)(p^2 - 5p - 6)} \\ &= \frac{-p^2 + 4p + 13}{(p + 2)(p + 1)(p - 6)} \end{aligned}$$ $$\begin{aligned} Y(p) &= \frac{A}{(p + 2)} + \frac{B}{(p + 1)} + \frac{C}{(p - 6)} \\ &= \frac{(A + B + C)p^2 + (-5A - 4B + 3C)p + (-6A - 12B + 2C)}{(p + 2)(p + 1)(p - 6)} \\ &\quad \left\{ \begin{array}{c} A + B + C = -1 \\ -5A - 4B + 3C = 4 \\ -6A - 12B + 2C = 13 \end{array} \right. \to \left\{ \begin{array}{l} A = \frac{1}{8} \\ B = -\frac{8}{7} \\ C = \frac{1}{56} \end{array} \right. \end{aligned}$$ $$Y(p) = \frac{1}{8} \cdot \frac{1}{(p + 2)} - \frac{8}{7} \cdot \frac{1}{(p + 1)} + \frac{1}{56} \cdot \frac{1}{(p - 6)} \leftarrow \frac{1}{8} e^{-2t} - \frac{8}{7} e^{-t} + \frac{1}{56} e^{6t} = y(t)$$

**Answer:** $y(t) = \frac{1}{8} e^{-2t} - \frac{8}{7} e^{-t} + \frac{1}{56} e^{6t}$.