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avatar+248 

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

Current answer (just need help with last part):

Let Y(s)=L[y(t)] :

--> [(s^2)Y(s) - sy(0) - y'(0)] - 2[sY(s) - y(0)] -3Y(s) = 0

--> [(s^2)Y(s) - s(1) - 0]        - 2[sY(s) - 1]     -3Y(s) = 0

                                                                   -->Y(s) = (s-2) / (s-3)(s+1)

 

This is where I'm stuck, I need to do the inverse laplace transform here to get the final answer but I'm not sure how I can do it with an equation of this form. Any help is appreciated.

difficulty advanced

Best Answer 

 #3
avatar+20035 
+20

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:

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

heureka  May 4, 2015
 #1
avatar+27061 
+15

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?

.

Alan  May 4, 2015
 #2
avatar+248 
+5

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

 #3
avatar+20035 
+20
Best Answer

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:

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

heureka  May 4, 2015
 #4
avatar+90088 
0

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

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

 

  

CPhill  May 4, 2015

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