We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
898
4
avatar+249 

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
 May 4, 2015

Best Answer 

 #3
avatar+22358 
+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:

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

.
 May 4, 2015
 #1
avatar+28029 
+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?

.

 May 4, 2015
 #2
avatar+249 
+5

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

 May 4, 2015
 #3
avatar+22358 
+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+101424 
0

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

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

 

  

 May 4, 2015

15 Online Users

avatar
avatar
avatar