Laplace Transforms -> Unsure how to do the inverse transform

Question: Solve y''(t) - 2y'(t)-3y(t) = 0 where y(0) = 1 and y'(0) = 0.

$$\boxed{ \begin{array}{rcl} \mathcal{L} \{ y''(t)\} -2\cdot \mathcal{L} \{ y'(t)\}-3\cdot \mathcal{L} \{ y(t)\} = \mathcal{L} \{0\}\\ \end{array} }$$

$$\boxed{ \begin{array}{rcl} \mathcal{L} \{ y''(t)\} &=& s^2\cdot \mathcal{L} \{ y(t)\} -s\cdot y(0)-y'(0)\\ \mathcal{L} \{ y''(t)\} &=& s^2\cdot \mathcal{L} \{ y(t)\} -s\cdot 1-0\\ \mathcal{L} \{ y''(t)\} &=& s^2\cdot \mathcal{L} \{ y(t)\} -s \\ \end{array} }$$

$$\boxed{ \begin{array}{rcl} -2\cdot \mathcal{L} \{ y'(t)\} &=& -2 \left[ s\cdot \mathcal{L} \{ y(t)\} -y(0)\right]\\ -2\cdot \mathcal{L} \{ y'(t)\} &=& -2 \left[ s\cdot \mathcal{L} \{ y(t)\} -1\right]\\ \end{array} }$$

$$\begin{array}{rcl} s^2\cdot \mathcal{L} \{ y(t)\} -s -2 \left[ s\cdot \mathcal{L} \{ y(t)\} -1\right] -3\cdot \mathcal{L} \{ y(t)\} &=& \mathcal{L} \{0\} \\ \mathcal{L} \{ y(t)\} \left[ s^2-2\cdot s -3 \right] &=& s-2\\ \mathcal{L} \{ y(t)\} \left[ (s+1)\cdot(s-3) \right] &=& s-2\\ \mathcal{L} \{ y(t)\} &=& \dfrac{ s-2 } {(s+1)\cdot(s-3)} \\ \end{array}$$

$$\boxed{ \begin{array}{rclcc} & \dfrac{ s-2 } {(s+1)\cdot(s-3)} &=& \dfrac{A}{s+1} + \dfrac{B}{s-3}\\\\ &s-2 &=& A \cdot (s-3) + B \cdot (s+1) \\\\ s=3: & 3-2 &=& 0 + B\cdot 4 \Rightarrow B = \dfrac{1}{4} \\ s=-1: & -1-2 &=& A\cdot(-4) + 0 \Rightarrow A = \dfrac{3}{4} \end{array} }$$

$$\begin{array}{rcl} \mathcal{L} \{ y(t)\} &=& \dfrac{ s-2 } {(s+1)\cdot(s-3)}=\dfrac{A}{s+1} + \dfrac{B}{s-3} = \dfrac{3}{4}\cdot \left( \dfrac{1}{s+1} \right) + \dfrac{1}{4} \cdot \left( \dfrac{1}{s-3} \right) \\\\ \mathcal{L} \{ y(t)\} &=& \dfrac{3}{4}\cdot \left( \dfrac{1}{s+1} \right) + \dfrac{1}{4} \cdot \left( \dfrac{1}{s-3} \right) \end{array}$$

$$\boxed{ \begin{array}{rcl} \mathcal{L}^{-1} \{ \dfrac{1}{s-a} \} &=& e^{a\cdot t} \\ \end{array} }$$

$$\textcolor[rgb]{150,0,0}{ \begin{array}{rcl} y(t) &=& \dfrac{3}{4}\cdot e^{-t} + \dfrac{1}{4} \cdot e^{3\cdot t} \end{array} }$$

Check:

$$\begin{array}{lrcl} (1)&: y(0) &=& \dfrac{3}{4}+\dfrac{1}{4} = 1 \qquad \text{okay} \\\\ (2)&: y'(t) &=& -\dfrac{3}{4}\cdot e^{-t} + 3\cdot \dfrac{1}{4} \cdot e^{3\cdot t}} \\\\ &: y'(0) &=& -\dfrac{3}{4} + \dfrac{3}{4} = 0 \qquad \text{okay}\\ \end{array}\\\\\\ \underbrace{ \begin{array}{lrcl} &y''(t) &=& \dfrac{3}{4}\cdot e^{-t} + \dfrac{9}{4}\cdot e^{3\cdot t} \\\\ & -2\cdot y'(t) & =& \dfrac{6}{4}\cdot e^{-t} -\dfrac{6}{4}\cdot e^ {3\cdot t} \\\\ & -3\cdot y(t) & =& -\dfrac{9}{4}\cdot e^{-t} -\dfrac{3}{4}\cdot e^{3\cdot t} \end{array} }_{ y''(t)-2\cdot y'(t)-3\cdot y(t)= 0 \qquad \text{okay} }$$

This is where partial fractions come in handy.

$$\frac{s-2}{(s-3)(s+1)}=\frac{3}{4(s+1)}+\frac{1}{4(s-3)}$$

Can you take it from here?

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Cheers yeah its straightforward now --> y(t) = (e^3t)/4 + (3e^-t)/4

Vey nice, Alan and heureka....!!!!

These transforms make those DEs a lot more simple to solve.....!!!!