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2.10 HW8

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2.10.1 Section 6.1 problem 7

Find Laplace Transform of $f\left ( t\right ) =\cosh \left ( bt\right )$

solution Since $\cosh \left ( bt\right ) =\frac{e^{bt}+e^{-bt}}{2}$ then $$\begin{align*} \mathcal{L}\cosh \left ( bt\right ) & =\frac{1}{2}\mathcal{L}\left ( e^{bt}+e^{-bt}\right ) \\ & =\frac{1}{2}\left ( \mathcal{L}e^{bt}+\mathcal{L}e^{-bt}\right ) \end{align*}$$

But $$\mathcal{L}e^{bt}=\frac{1}{s-b}$$

For $s>b$ and

$$\mathcal{L}e^{bt}=\frac{1}{s-b}$$

For $s<b$. Hence

$$\begin{align*} \mathcal{L}\cosh \left ( bt\right ) & =\frac{1}{2}\left ( \frac{1}{s-b}+\frac{1}{s+b}\right ) \\ & =\frac{s^{2}}{s^{2}-b^{2}} \end{align*}$$

For $s>\left \vert b\right \vert$

2.10.2 Section 6.1 problem 8

Find Laplace Transform of $f\left ( t\right ) =\sinh \left ( bt\right )$

solution Since $\sinh \left ( bt\right ) =\frac{e^{bt}-e^{-bt}}{2}$ then $$\begin{align*} \mathcal{L}\sinh \left ( bt\right ) & =\frac{1}{2}\mathcal{L}\left ( e^{bt}-e^{-bt}\right ) \\ & =\frac{1}{2}\left ( \mathcal{L}e^{bt}-\mathcal{L}e^{-bt}\right ) \end{align*}$$

But, as we found in the last problem $$\mathcal{L}e^{bt}=\frac{1}{s-b}\qquad s>b$$ And $$\mathcal{L}e^{-bt}=\frac{1}{s+b}\qquad s<b$$ Therefore $$\begin{align*} \mathcal{L}\sinh \left ( bt\right ) & =\frac{1}{2}\left ( \frac{1}{s-b}-\frac{1}{s+b}\right ) \qquad s>b;s<b\\ & =\frac{b}{s^{2}-b^{2}}\qquad s>\left \vert b\right \vert \end{align*}$$

2.10.3 Section 6.1 problem 9

Find Laplace Transform of $f\left ( t\right ) =e^{at}\cosh \left ( bt\right )$

solution Using the property that $$e^{at}f\left ( t\right ) \Longleftrightarrow F\left ( s-a\right )$$ Where $f\left ( t\right ) =\cosh \left ( bt\right )$ now. We already found above that $\cosh \left ( bt\right ) \Longleftrightarrow \frac{s}{s^{2}-b^{2}}$, for $s>\left \vert b\right \vert$. In other words, $F\left ( s\right ) =\frac{s}{s^{2}-b^{2}}$, therefore $$e^{at}\cosh \left ( bt\right ) \Longleftrightarrow \frac{\left ( s-a\right ) }{\left ( s-a\right ) ^{2}-b^{2}}\qquad s-a>\left \vert b\right \vert$$

2.10.4 Section 6.1 problem 10

Find Laplace Transform of $f\left ( t\right ) =e^{at}\sinh \left ( bt\right )$

solution Using the property that $$e^{at}f\left ( t\right ) \Longleftrightarrow F\left ( s-a\right )$$ Where $f\left ( t\right ) =\sinh \left ( bt\right )$ now. We already found above that $\sinh \left ( bt\right ) \Longleftrightarrow \frac{b}{s^{2}-b^{2}}$, for $s>\left \vert b\right \vert$. In other words, $F\left ( s\right ) =\frac{b}{s^{2}-b^{2}}$, therefore $$e^{at}\sinh \left ( bt\right ) \Longleftrightarrow \frac{b}{\left ( s-a\right ) ^{2}-b^{2}}\qquad s-a>\left \vert b\right \vert$$

2.10.5 Section 6.2 problem 17

Use Laplace transform to solve $y^{\left ( 4\right ) }-4y^{\prime \prime \prime }+6y^{\prime \prime }-4y^{\prime }+y=0$ for $y\left ( 0\right ) =0,y^{\prime }\left ( 0\right ) =1,y^{\prime \prime }\left ( 0\right ) =0,y^{\prime \prime \prime }\left ( 0\right ) =1$

Solution Taking Laplace transform of the ODE gives

$$\begin{equation} \mathcal{L}\left \{ y^{\left ( 4\right ) }\right \} -4\mathcal{L}\left \{ y^{\prime \prime \prime }\right \} +6\mathcal{L}\left \{ y^{\prime \prime }\right \} -4\mathcal{L}\left \{ y^{\prime }\right \} +\mathcal{L}\left \{ y\right \} =0 \tag{1} \end{equation}$$ Let $\mathcal{L}\left \{ y\right \} =Y\left ( s\right )$ then $$\begin{align*} \mathcal{L}\left \{ y^{\left ( 4\right ) }\right \} & =s^{4}Y\left ( s\right ) -s^{3}y\left ( 0\right ) -s^{2}y^{\prime }\left ( 0\right ) -sy^{\prime \prime }\left ( 0\right ) -y^{\prime \prime \prime }\left ( 0\right ) \\ & =s^{4}Y\left ( s\right ) -s^{3}\left ( 0\right ) -s^{2}\left ( 1\right ) -s\left ( 0\right ) -1\\ & =s^{4}Y\left ( s\right ) -s^{2}-1 \end{align*}$$

And $$\begin{align*} \mathcal{L}\left \{ y^{\prime \prime \prime }\right \} & =s^{3}Y\left ( s\right ) -s^{2}y\left ( 0\right ) -sy^{\prime }\left ( 0\right ) -y^{\prime \prime }\left ( 0\right ) \\ & =s^{3}Y\left ( s\right ) -s^{2}\left ( 0\right ) -s\left ( 1\right ) -0\\ & =s^{3}Y\left ( s\right ) -s \end{align*}$$

And $$\begin{align*} \mathcal{L}\left \{ y^{\prime \prime }\right \} & =s^{2}Y\left ( s\right ) -sy\left ( 0\right ) -y^{\prime }\left ( 0\right ) \\ & =s^{2}Y\left ( s\right ) -s\left ( 0\right ) -1\\ & =s^{2}Y\left ( s\right ) -1 \end{align*}$$

And $$\begin{align*} \mathcal{L}\left \{ y^{\prime }\right \} & =sY\left ( s\right ) -y\left ( 0\right ) \\ & =sY\left ( s\right ) \end{align*}$$

Hence (1) becomes $$\begin{align*} \left ( s^{4}Y\left ( s\right ) -s^{2}-1\right ) -4\left ( s^{3}Y\left ( s\right ) -s\right ) +6\left ( s^{2}Y\left ( s\right ) -1\right ) -4\left ( sY\left ( s\right ) \right ) +Y\left ( s\right ) & =0\\ Y\left ( s\right ) \left ( s^{4}-4s^{3}+6s^{2}-4s+1\right ) -s^{2}-1+4s-6 & =0 \end{align*}$$

Therefore $$\begin{align} Y\left ( s\right ) & =\frac{s^{2}-4s+7}{s^{4}-4s^{3}+6s^{2}-4s+1}\nonumber \\ & =\frac{s^{2}-4s+7}{\left ( s-1\right ) ^{4}}\nonumber \\ & =\frac{s^{2}}{\left ( s-1\right ) ^{4}}-\frac{4s}{\left ( s-1\right ) ^{4}}+\frac{7}{\left ( s-1\right ) ^{4}} \tag{2} \end{align}$$

But $$\begin{align*} \frac{s^{2}}{\left ( s-1\right ) ^{4}} & =\frac{\left ( s-1\right ) ^{2}-1+2s}{\left ( s-1\right ) ^{4}}\\ & =\frac{\left ( s-1\right ) ^{2}}{\left ( s-1\right ) ^{4}}-\frac{1}{\left ( s-1\right ) ^{4}}+2\frac{\left ( s-1\right ) +1}{\left ( s-1\right ) ^{4}}\\ & =\frac{1}{\left ( s-1\right ) ^{2}}-\frac{1}{\left ( s-1\right ) ^{4}}+2\frac{\left ( s-1\right ) }{\left ( s-1\right ) ^{4}}+2\frac{1}{\left ( s-1\right ) ^{4}}\\ & =\frac{1}{\left ( s-1\right ) ^{2}}-\frac{1}{\left ( s-1\right ) ^{4}}+2\frac{1}{\left ( s-1\right ) ^{3}}+2\frac{1}{\left ( s-1\right ) ^{4}}\\ & =\frac{1}{\left ( s-1\right ) ^{2}}+\frac{2}{\left ( s-1\right ) ^{3}}+\frac{1}{\left ( s-1\right ) ^{4}} \end{align*}$$

And $$\begin{align*} \frac{4s}{\left ( s-1\right ) ^{4}} & =4\frac{\left ( s-1\right ) +1}{\left ( s-1\right ) ^{4}}\\ & =4\frac{\left ( s-1\right ) }{\left ( s-1\right ) ^{4}}+4\frac{1}{\left ( s-1\right ) ^{4}}\\ & =\frac{4}{\left ( s-1\right ) ^{3}}+\frac{4}{\left ( s-1\right ) ^{4}} \end{align*}$$

Therefore (2) becomes $$\begin{align} Y\left ( s\right ) & =\left ( \frac{1}{\left ( s-1\right ) ^{2}}+\frac{2}{\left ( s-1\right ) ^{3}}+\frac{1}{\left ( s-1\right ) ^{4}}\right ) -\left ( \frac{4}{\left ( s-1\right ) ^{3}}+\frac{4}{\left ( s-1\right ) ^{4}}\right ) +\frac{7}{\left ( s-1\right ) ^{4}}\nonumber \\ & =\frac{1}{\left ( s-1\right ) ^{2}}-\frac{2}{\left ( s-1\right ) ^{3}}+\frac{4}{\left ( s-1\right ) ^{4}} \tag{3} \end{align}$$

Now using property the shift property of $F\left ( s\right )$ together with $$\begin{align*} \frac{1}{s^{2}} & \Longleftrightarrow t\\ \frac{1}{s^{3}} & \Longleftrightarrow \frac{t^{2}}{2}\\ \frac{1}{s^{4}} & \Longleftrightarrow \frac{t^{3}}{6} \end{align*}$$

Therefore $$\begin{align*} \frac{1}{\left ( s-1\right ) ^{2}} & \Longleftrightarrow e^{t}t\\ \frac{1}{\left ( s-1\right ) ^{3}} & \Longleftrightarrow e^{t}\frac{t^{2}}{2}\\ \frac{1}{\left ( s-1\right ) ^{4}} & \Longleftrightarrow e^{t}\frac{t^{3}}{6} \end{align*}$$

And (3) becomes $$\begin{align*} \frac{1}{\left ( s-1\right ) ^{2}}-\frac{2}{\left ( s-1\right ) ^{3}}+\frac{4}{\left ( s-1\right ) ^{4}} & \Longleftrightarrow e^{t}t-2\left ( e^{t}\frac{t^{2}}{2}\right ) +4\left ( e^{t}\frac{t^{3}}{6}\right ) \\ & =e^{t}t-e^{t}t^{2}+\frac{2}{3}e^{t}t^{3} \end{align*}$$

Hence $$y\left ( t\right ) =e^{t}\left ( t-t^{2}+\frac{2}{3}t^{3}\right )$$

2.10.6 Section 6.2 problem 18

Use Laplace transform to solve $y^{\left ( 4\right ) }-y=0$ for $y\left ( 0\right ) =1,y^{\prime }\left ( 0\right ) =0,y^{\prime \prime }\left ( 0\right ) =1,y^{\prime \prime \prime }\left ( 0\right ) =0$

Solution Taking Laplace transform of the ODE gives $$\begin{equation} \mathcal{L}\left \{ y^{\left ( 4\right ) }\right \} -\mathcal{L}\left \{ y\right \} =0 \tag{1} \end{equation}$$ Let $\mathcal{L}\left \{ y\right \} =Y\left ( s\right )$ then $$\begin{align*} \mathcal{L}\left \{ y^{\left ( 4\right ) }\right \} & =s^{4}Y\left ( s\right ) -s^{3}y\left ( 0\right ) -s^{2}y^{\prime }\left ( 0\right ) -sy^{\prime \prime }\left ( 0\right ) -y^{\prime \prime \prime }\left ( 0\right ) \\ & =s^{4}Y\left ( s\right ) -s^{3}\left ( 1\right ) -s^{2}\left ( 0\right ) -s\left ( 1\right ) -0\\ & =s^{4}Y\left ( s\right ) -s^{3}-s \end{align*}$$

Hence (1) becomes $$s^{4}Y\left ( s\right ) -s^{3}-s-Y\left ( s\right ) =0$$ Solving for $Y\left ( s\right )$ gives $$\begin{align*} Y\left ( s\right ) & =\frac{s^{3}+s}{s^{4}-1}\\ & =\frac{s\left ( s^{2}+1\right ) }{s^{4}-1}\\ & =\frac{s\left ( s^{2}+1\right ) }{\left ( s^{2}-1\right ) \left ( s^{2}+1\right ) }\\ & =\frac{s}{s^{2}-1} \end{align*}$$

But, Hence above becomes, where $a=1$ $$\frac{s}{s^{2}-1}\Longleftrightarrow \cosh \left ( t\right )$$ Hence $$y\left ( t\right ) =\cosh \left ( at\right )$$

2.10.7 Section 6.2 problem 19

Use Laplace transform to solve $y^{\left ( 4\right ) }-4y=0$ for $y\left ( 0\right ) =1,y^{\prime }\left ( 0\right ) =0,y^{\prime \prime }\left ( 0\right ) =-2,y^{\prime \prime \prime }\left ( 0\right ) =0$

Solution Taking Laplace transform of the ODE gives $$\begin{equation} \mathcal{L}\left \{ y^{\left ( 4\right ) }\right \} -4\mathcal{L}\left \{ y\right \} =0 \tag{1} \end{equation}$$ Let $\mathcal{L}\left \{ y\right \} =Y\left ( s\right )$ then $$\begin{align*} \mathcal{L}\left \{ y^{\left ( 4\right ) }\right \} & =s^{4}Y\left ( s\right ) -s^{3}y\left ( 0\right ) -s^{2}y^{\prime }\left ( 0\right ) -sy^{\prime \prime }\left ( 0\right ) -y^{\prime \prime \prime }\left ( 0\right ) \\ & =s^{4}Y\left ( s\right ) -s^{3}\left ( 1\right ) -s^{2}\left ( 0\right ) -s\left ( -2\right ) -0\\ & =s^{4}Y\left ( s\right ) -s^{3}+2s \end{align*}$$

Hence (1) becomes $$s^{4}Y\left ( s\right ) -s^{3}+2s-4Y\left ( s\right ) =0$$ Solving for $Y\left ( s\right )$ gives $$\begin{align*} Y\left ( s\right ) & =\frac{s^{3}-2s}{s^{4}-4}\\ & =\frac{s^{3}-2s}{\left ( s^{2}-2\right ) \left ( s^{2}+2\right ) }\\ & =\frac{s\left ( s^{2}-2\right ) }{\left ( s^{2}-2\right ) \left ( s^{2}+2\right ) }\\ & =\frac{s}{\left ( s^{2}+2\right ) } \end{align*}$$

Using $\cos \left ( at\right ) \Longleftrightarrow \frac{s}{s^{2}+a^{2}},$ the above becomes, where $a=\sqrt{2}$ $$\frac{s}{\left ( s^{2}+2\right ) }\Longleftrightarrow \cos \left ( \sqrt{2}t\right )$$ Hence $$y\left ( t\right ) =\cos \left ( \sqrt{2}t\right )$$

2.10.8 Section 6.2 problem 20

Use Laplace transform to solve $y^{\prime \prime }+\omega ^{2}y=\cos 2t;$ $\omega ^{2}\neq 4;$ $y\left ( 0\right ) =1,y^{\prime }\left ( 0\right ) =0$

Solution Let $Y\left ( s\right ) =\mathcal{L}\left \{ y\left ( t\right ) \right \}$. Taking Laplace transform of the ODE, and using $\cos \left ( at\right ) \Longleftrightarrow \frac{s}{s^{2}+a^{2}}$ gives $$\begin{equation} s^{2}Y\left ( s\right ) -sy\left ( 0\right ) -y^{\prime }\left ( 0\right ) +\omega ^{2}Y\left ( s\right ) =\frac{s}{s^{2}+4} \tag{1} \end{equation}$$ Applying initial conditions $$s^{2}Y\left ( s\right ) -s+\omega ^{2}Y\left ( s\right ) =\frac{s}{s^{2}+4}$$ Solving for $Y\left ( s\right )$ $$\begin{align} Y\left ( s\right ) \left ( s^{2}+\omega ^{2}\right ) -s & =\frac{s}{s^{2}+4}\nonumber \\ Y\left ( s\right ) & =\frac{s}{\left ( s^{2}+4\right ) \left ( s^{2}+\omega ^{2}\right ) }+\frac{s}{\left ( s^{2}+\omega ^{2}\right ) } \tag{2} \end{align}$$

But $$\begin{align*} \frac{s}{\left ( s^{2}+4\right ) \left ( s^{2}+\omega ^{2}\right ) } & =\frac{As+B}{\left ( s^{2}+4\right ) }+\frac{Cs+D}{\left ( s^{2}+\omega ^{2}\right ) }\\ s & =\left ( As+B\right ) \left ( s^{2}+\omega ^{2}\right ) +\left ( Cs+D\right ) \left ( s^{2}+4\right ) \\ s & =4D+As^{3}+Bs^{2}+Cs^{3}+B\omega ^{2}+s^{2}D+4Cs+As\omega ^{2}\allowbreak \\ s & =\left ( 4D+B\omega ^{2}\right ) +s\left ( 4C+A\omega ^{2}\right ) +s^{2}\left ( B+D\right ) +s^{3}\left ( A+C\right ) \end{align*}$$

Hence $$\begin{align*} 4D+B\omega ^{2} & =0\\ 4C+A\omega ^{2} & =1\\ B+D & =0\\ A+C & =0 \end{align*}$$

Equation (2,4) gives  $A=\frac{1}{\omega ^{2}-4},C=\frac{1}{4-\omega ^{2}}$ and (1,3) gives $B=0,D=0$. Hence $$\frac{s}{\left ( s^{2}+4\right ) \left ( s^{2}+\omega ^{2}\right ) }=\left ( \frac{1}{\omega ^{2}-4}\right ) \frac{s}{\left ( s^{2}+4\right ) }+\left ( \frac{1}{4-\omega ^{2}}\right ) \frac{s}{\left ( s^{2}+\omega ^{2}\right ) }$$ Therefore (2) becomes $$\begin{align*} Y\left ( s\right ) & =\left ( \frac{1}{\omega ^{2}-4}\right ) \frac{s}{\left ( s^{2}+4\right ) }+\left ( \frac{1}{4-\omega ^{2}}\right ) \frac{s}{\left ( s^{2}+\omega ^{2}\right ) }+\frac{s}{\left ( s^{2}+\omega ^{2}\right ) }\\ & =\left ( \frac{1}{\omega ^{2}-4}\right ) \frac{s}{\left ( s^{2}+4\right ) }+\left ( \frac{5-\omega ^{2}}{4-\omega ^{2}}\right ) \frac{s}{\left ( s^{2}+\omega ^{2}\right ) } \end{align*}$$

Using $\cos \left ( at\right ) \Longleftrightarrow \frac{s}{s^{2}+a^{2}}$, the above becomes $$\begin{align*} \left ( \frac{1}{\omega ^{2}-4}\right ) \frac{s}{\left ( s^{2}+4\right ) }+\left ( \frac{5-\omega ^{2}}{4-\omega ^{2}}\right ) \frac{s}{\left ( s^{2}+\omega ^{2}\right ) } & \Longleftrightarrow \left ( \frac{1}{\omega ^{2}-4}\right ) \cos \left ( 2t\right ) +\left ( \frac{5-\omega ^{2}}{4-\omega ^{2}}\right ) \cos \left ( \omega t\right ) \\ & =\left ( \frac{1}{\omega ^{2}-4}\right ) \cos \left ( 2t\right ) +\left ( \frac{\omega ^{2}-5}{\omega ^{2}-4}\right ) \cos \left ( \omega t\right ) \end{align*}$$

Hence $$\begin{align*} y\left ( t\right ) & =\left ( \frac{1}{\omega ^{2}-4}\right ) \cos \left ( 2t\right ) +\left ( \frac{\omega ^{2}-5}{\omega ^{2}-4}\right ) \cos \left ( \omega t\right ) \\ & =\frac{\left ( \omega ^{2}-5\right ) \cos \left ( \omega t\right ) +\cos \left ( 2t\right ) }{\omega ^{2}-4} \end{align*}$$

2.10.9 Section 6.2 problem 21

Use Laplace transform to solve $y^{\prime \prime }-2y^{\prime }+2y=\cos t;$ $y\left ( 0\right ) =1,y^{\prime }\left ( 0\right ) =0$

Solution Let $Y\left ( s\right ) =\mathcal{L}\left \{ y\left ( t\right ) \right \}$. Taking Laplace transform of the ODE, and using $\cos \left ( at\right ) \Longleftrightarrow \frac{s}{s^{2}+a^{2}}$ gives $$\begin{equation} \left ( s^{2}Y\left ( s\right ) -sy\left ( 0\right ) -y^{\prime }\left ( 0\right ) \right ) -2\left ( sY\left ( s\right ) -y\left ( 0\right ) \right ) +2Y\left ( s\right ) =\frac{s}{s^{2}+1} \tag{1} \end{equation}$$ Applying initial conditions $$s^{2}Y\left ( s\right ) -s-2\left ( sY\left ( s\right ) -1\right ) +2Y\left ( s\right ) =\frac{s}{s^{2}+1}$$ Solving for $Y\left ( s\right )$ $$\begin{align} s^{2}Y\left ( s\right ) -s-2sY\left ( s\right ) +2+2Y\left ( s\right ) & =\frac{s}{s^{2}+1}\nonumber \\ Y\left ( s\right ) \left ( s^{2}-2s+2\right ) -s+2 & =\frac{s}{s^{2}+1}\nonumber \\ Y\left ( s\right ) & =\frac{s}{\left ( s^{2}+1\right ) \left ( s^{2}-2s+2\right ) }+\frac{s}{\left ( s^{2}-2s+2\right ) }-\frac{2}{\left ( s^{2}-2s+2\right ) } \tag{2} \end{align}$$

But $$\begin{align*} \frac{s}{\left ( s^{2}+1\right ) \left ( s^{2}-2s+2\right ) } & =\frac{As+B}{\left ( s^{2}+1\right ) }+\frac{Cs+D}{s^{2}-2s+2}\\ s & =\left ( As+B\right ) \left ( s^{2}-2s+2\right ) +\left ( Cs+D\right ) \left ( s^{2}+1\right ) \\ s & =2B+D-2As^{2}+As^{3}+Bs^{2}+Cs^{3}+s^{2}D+2As-\allowbreak 2Bs+Cs\\ s & =\left ( 2B+D\right ) +s\left ( 2A-2B+C\right ) +s^{2}\left ( -2A+B+D\right ) +s^{3}\left ( A+C\right ) \end{align*}$$

Hence $$\begin{align*} 2B+D & =0\\ 2A-2B+C & =1\\ -2A+B+D & =0\\ A+C & =0 \end{align*}$$

Solving gives $A=\frac{1}{5},B=-\frac{2}{5},C=-\frac{1}{5},D=\frac{4}{5}$, hence $$\begin{align} \frac{s}{\left ( s^{2}+1\right ) \left ( s^{2}-2s+2\right ) } & =\frac{1}{5}\frac{s-2}{\left ( s^{2}+1\right ) }-\frac{1}{5}\frac{s-4}{s^{2}-2s+2}\nonumber \\ & =\frac{1}{5}\frac{s}{s^{2}+1}-\frac{2}{5}\frac{1}{s^{2}+1}-\frac{1}{5}\frac{s}{s^{2}-2s+2}+\frac{4}{5}\frac{1}{s^{2}-2s+2} \tag{3} \end{align}$$

Completing the squares for $$\begin{align*} s^{2}-2s+2 & =a\left ( s+b\right ) ^{2}+d\\ & =a\left ( s^{2}+b^{2}+2bs\right ) +d\\ & =as^{2}+ab^{2}+2abs+d \end{align*}$$

Hence $a=1,2ab=-2,\left ( ab^{2}+d\right ) =2$, hence $b=-1,d=1$, hence $$s^{2}-2s+2=\left ( s-1\right ) ^{2}+1$$ Hence (3) becomes $$\begin{align*} \frac{s}{\left ( s^{2}+1\right ) \left ( s^{2}-2s+2\right ) } & =\frac{1}{5}\frac{s}{s^{2}+1}-\frac{2}{5}\frac{1}{s^{2}+1}-\frac{1}{5}\frac{s}{\left ( s-1\right ) ^{2}+1}+\frac{4}{5}\frac{1}{\left ( s-1\right ) ^{2}+1}\\ & =\frac{1}{5}\frac{s}{s^{2}+1}-\frac{2}{5}\frac{1}{s^{2}+1}-\frac{1}{5}\frac{\left ( s-1\right ) +1}{\left ( s-1\right ) ^{2}+1}+\frac{4}{5}\frac{1}{\left ( s-1\right ) ^{2}+1}\\ & =\frac{1}{5}\frac{s}{s^{2}+1}-\frac{2}{5}\frac{1}{s^{2}+1}-\frac{1}{5}\frac{\left ( s-1\right ) }{\left ( s-1\right ) ^{2}+1}-\frac{1}{5}\frac{1}{\left ( s-1\right ) ^{2}+1}+\frac{4}{5}\frac{1}{\left ( s-1\right ) ^{2}+1}\\ & =\frac{1}{5}\frac{s}{s^{2}+1}-\frac{2}{5}\frac{1}{s^{2}+1}-\frac{1}{5}\frac{\left ( s-1\right ) }{\left ( s-1\right ) ^{2}+1}+\frac{3}{5}\frac{1}{\left ( s-1\right ) ^{2}+1} \end{align*}$$

Therefore (2) becomes $$\begin{align*} Y\left ( s\right ) & =\frac{1}{5}\frac{s}{s^{2}+1}-\frac{2}{5}\frac{1}{s^{2}+1}-\frac{1}{5}\frac{\left ( s-1\right ) }{\left ( s-1\right ) ^{2}+1}+\frac{3}{5}\frac{1}{\left ( s-1\right ) ^{2}+1}+\frac{s}{\left ( s-1\right ) ^{2}+1}-\frac{2}{\left ( s-1\right ) ^{2}+1}\\ & =\frac{1}{5}\frac{s}{s^{2}+1}-\frac{2}{5}\frac{1}{s^{2}+1}-\frac{1}{5}\frac{\left ( s-1\right ) }{\left ( s-1\right ) ^{2}+1}+\frac{3}{5}\frac{1}{\left ( s-1\right ) ^{2}+1}+\frac{\left ( s-1\right ) +1}{\left ( s-1\right ) ^{2}+1}-\frac{2}{\left ( s-1\right ) ^{2}+1}\\ & =\frac{1}{5}\frac{s}{s^{2}+1}-\frac{2}{5}\frac{1}{s^{2}+1}-\frac{1}{5}\frac{\left ( s-1\right ) }{\left ( s-1\right ) ^{2}+1}+\frac{3}{5}\frac{1}{\left ( s-1\right ) ^{2}+1}+\frac{\left ( s-1\right ) }{\left ( s-1\right ) ^{2}+1}+\frac{1}{\left ( s-1\right ) ^{2}+1}-\frac{2}{\left ( s-1\right ) ^{2}+1}\\ & =\frac{1}{5}\frac{s}{s^{2}+1}-\frac{2}{5}\frac{1}{s^{2}+1}+\frac{4}{5}\frac{\left ( s-1\right ) }{\left ( s-1\right ) ^{2}+1}-\frac{2}{5}\frac{1}{\left ( s-1\right ) ^{2}+1} \end{align*}$$

Using $\cos \left ( at\right ) \Longleftrightarrow \frac{s}{s^{2}+a^{2}},\sin \left ( at\right ) \Longleftrightarrow \frac{a}{s^{2}+a^{2}}$ and the shift property of Laplace transform, then $$\begin{align*} \frac{1}{5}\frac{s}{s^{2}+1} & \Longleftrightarrow \frac{1}{5}\cos \left ( t\right ) \\ \frac{2}{5}\frac{1}{s^{2}+1} & \Longleftrightarrow \frac{2}{5}\sin \left ( t\right ) \\ \frac{4}{5}\frac{\left ( s-1\right ) }{\left ( s-1\right ) ^{2}+1} & \Longleftrightarrow \frac{4}{5}e^{t}\cos t\\ \frac{2}{5}\frac{1}{\left ( s-1\right ) ^{2}+1} & \Longleftrightarrow \frac{8}{5}e^{t}\sin t \end{align*}$$

Hence $$\begin{align*} y\left ( t\right ) & =\frac{1}{5}\cos \left ( t\right ) -\frac{2}{5}\sin \left ( t\right ) +\frac{4}{5}e^{t}\cos t-\frac{2}{5}e^{t}\sin t\\ & \frac{1}{5}\left ( \cos t-2\sin t+4e^{t}\cos t-2e^{t}\sin t\right ) \end{align*}$$

2.10.10 Section 6.2 problem 22

Use Laplace transform to solve $y^{\prime \prime }-2y^{\prime }+2y=e^{-t};y\left ( 0\right ) =0,y^{\prime }\left ( 0\right ) =1$

Solution Let $Y\left ( s\right ) =\mathcal{L}\left \{ y\left ( t\right ) \right \}$. Taking Laplace transform of the ODE, and using $e^{-t}\Longleftrightarrow \frac{1}{s+1}$ gives $$\begin{equation} \left ( s^{2}Y\left ( s\right ) -sy\left ( 0\right ) -y^{\prime }\left ( 0\right ) \right ) -2\left ( sY\left ( s\right ) -y\left ( 0\right ) \right ) +2Y\left ( s\right ) =\frac{1}{s+1} \tag{1} \end{equation}$$ Applying initial conditions gives $$s^{2}Y\left ( s\right ) -1-2sY\left ( s\right ) +2Y\left ( s\right ) =\frac{1}{s+1}$$ Solving for $Y\left ( s\right )$ $$\begin{align} Y\left ( s\right ) \left ( s^{2}-2s+2\right ) -1 & =\frac{1}{s+1}\nonumber \\ Y\left ( s\right ) & =\frac{1}{\left ( s+1\right ) \left ( s^{2}-2s+2\right ) }+\frac{1}{s^{2}-2s+2} \tag{2} \end{align}$$

But $$\begin{align*} \frac{1}{\left ( s+1\right ) \left ( s^{2}-2s+2\right ) } & =\frac{A}{s+1}+\frac{Bs+C}{s^{2}-2s+2}\\ 1 & =A\left ( s^{2}-2s+2\right ) +\left ( Bs+C\right ) \left ( s+1\right ) \\ 1 & =2A+C+As^{2}+Bs^{2}-2As+Bs+Cs\\ 1 & =\left ( 2A+C\right ) +s\left ( -2A+B+C\right ) +s^{2}\left ( A+B\right ) \end{align*}$$

Hence $$\begin{align*} 1 & =2A+C\\ 0 & =-2A+B+C\\ 0 & =A+B \end{align*}$$

Solving gives $A=\frac{1}{5},B=-\frac{1}{5},C=\frac{3}{5}$, therefore $$\begin{align*} \frac{1}{\left ( s+1\right ) \left ( s^{2}-2s+2\right ) } & =\frac{1}{5}\frac{1}{s+1}+\frac{-\frac{1}{5}s+\frac{3}{5}}{s^{2}-2s+2}\\ & =\frac{1}{5}\frac{1}{s+1}-\frac{1}{5}\frac{s}{s^{2}-2s+2}++\frac{3}{5}\frac{1}{s^{2}-2s+2} \end{align*}$$

Completing the square for $s^{2}-2s+2$ which was done in last problem, gives $\left ( s-1\right ) ^{2}+1$, hence the above becomes $$\begin{align*} \frac{1}{\left ( s+1\right ) \left ( s^{2}-2s+2\right ) } & =\frac{1}{5}\frac{1}{s+1}-\frac{1}{5}\frac{s}{\left ( s-1\right ) ^{2}+1}++\frac{3}{5}\frac{1}{\left ( s-1\right ) ^{2}+1}\\ & =\frac{1}{5}\frac{1}{s+1}-\frac{1}{5}\frac{\left ( s-1\right ) +1}{\left ( s-1\right ) ^{2}+1}+\frac{3}{5}\frac{1}{\left ( s-1\right ) ^{2}+1}\\ & =\frac{1}{5}\frac{1}{s+1}-\frac{1}{5}\frac{\left ( s-1\right ) }{\left ( s-1\right ) ^{2}+1}-\frac{1}{5}\frac{1}{\left ( s-1\right ) ^{2}+1}+\frac{3}{5}\frac{1}{\left ( s-1\right ) ^{2}+1}\\ & =\frac{1}{5}\frac{1}{s+1}-\frac{1}{5}\frac{\left ( s-1\right ) }{\left ( s-1\right ) ^{2}+1}+\frac{2}{5}\frac{1}{\left ( s-1\right ) ^{2}+1} \end{align*}$$

Therefore (2) becomes $$Y\left ( s\right ) =\frac{1}{5}\frac{1}{s+1}-\frac{1}{5}\frac{\left ( s-1\right ) }{\left ( s-1\right ) ^{2}+1}+\frac{2}{5}\frac{1}{\left ( s-1\right ) ^{2}+1}+\frac{1}{\left ( s-1\right ) ^{2}+1}$$ Using $\cos \left ( at\right ) \Longleftrightarrow \frac{s}{s^{2}+a^{2}},\sin \left ( at\right ) \Longleftrightarrow \frac{a}{s^{2}+a^{2}}$ and the shift property of Laplace transform, then $$\begin{align*} \frac{1}{5}\frac{1}{s+1} & \Longleftrightarrow \frac{1}{5}e^{-t}\\ \frac{1}{5}\frac{\left ( s-1\right ) }{\left ( s-1\right ) ^{2}+1} & \Longleftrightarrow \frac{1}{5}e^{t}\cos t\\ \frac{2}{5}\frac{1}{\left ( s-1\right ) ^{2}+1} & \Longleftrightarrow \frac{2}{5}e^{t}\sin t\\ \frac{1}{\left ( s-1\right ) ^{2}+1} & \Longleftrightarrow e^{t}\sin t \end{align*}$$

Hence $$\begin{align*} y\left ( t\right ) & =\frac{1}{5}e^{-t}-\frac{1}{5}e^{t}\cos t+\frac{2}{5}e^{t}\sin t+e^{t}\sin t\\ & =\frac{1}{5}\left ( e^{-t}-e^{t}\cos t+7e^{t}\sin t\right ) \end{align*}$$

2.10.11 Section 6.2 problem 23

Use Laplace transform to solve $y^{\prime \prime }+2y^{\prime }+y=4e^{-t};y\left ( 0\right ) =2,y^{\prime }\left ( 0\right ) =-1$

Solution Let $Y\left ( s\right ) =\mathcal{L}\left \{ y\left ( t\right ) \right \}$. Taking Laplace transform of the ODE, and using $e^{-t}\Longleftrightarrow \frac{1}{s+1}$ gives $$\begin{equation} \left ( s^{2}Y\left ( s\right ) -sy\left ( 0\right ) -y^{\prime }\left ( 0\right ) \right ) +2\left ( sY\left ( s\right ) -y\left ( 0\right ) \right ) +Y\left ( s\right ) =\frac{4}{s+1} \tag{1} \end{equation}$$ Applying initial conditions gives $$\left ( s^{2}Y\left ( s\right ) -2s+1\right ) +2\left ( sY\left ( s\right ) -2\right ) +Y\left ( s\right ) =\frac{4}{s+1}$$ Solving for $Y\left ( s\right )$ $$\begin{align*} Y\left ( s\right ) \left ( s^{2}+2s+1\right ) -2s+1-4 & =\frac{4}{s+1}\\ Y\left ( s\right ) \left ( s^{2}+2s+1\right ) & =\frac{4}{s+1}+2s-1+4\\ Y\left ( s\right ) & =\frac{4}{\left ( s+1\right ) \left ( s^{2}+2s+1\right ) }+\frac{2s}{\left ( s^{2}+2s+1\right ) }-\frac{1}{\left ( s^{2}+2s+1\right ) }+\frac{4}{\left ( s^{2}+2s+1\right ) } \end{align*}$$

But $\left ( s^{2}+2s+1\right ) =\left ( s+1\right ) ^{2}$, hence $$\begin{equation} Y\left ( s\right ) =\frac{4}{\left ( s+1\right ) ^{3}}+\frac{2s}{\left ( s+1\right ) ^{2}}-\frac{1}{\left ( s+1\right ) ^{2}}+\frac{4}{\left ( s+1\right ) ^{2}} \tag{2} \end{equation}$$ But $$\begin{align*} \frac{2s}{\left ( s+1\right ) ^{2}} & =2\frac{s+1-1}{\left ( s+1\right ) ^{2}}\\ & =2\frac{\left ( s+1\right ) }{\left ( s+1\right ) ^{2}}-2\frac{1}{\left ( s+1\right ) ^{2}}\\ & =2\frac{1}{s+1}-2\frac{1}{\left ( s+1\right ) ^{2}} \end{align*}$$

Hence (2) becomes $$\begin{equation} Y\left ( s\right ) =\frac{4}{\left ( s+1\right ) ^{3}}+2\frac{1}{s+1}-2\frac{1}{\left ( s+1\right ) ^{2}}-\frac{1}{\left ( s+1\right ) ^{2}}+\frac{4}{\left ( s+1\right ) ^{2}} \tag{3} \end{equation}$$ We now ready to do the inversion. Since $\frac{1}{s^{3}}\Longleftrightarrow \frac{t^{2}}{2}$ and $\frac{1}{s^{2}}\Longleftrightarrow t$ and $\frac{1}{s}\Longleftrightarrow 1$ and using the shift property $e^{at}f\left ( t\right ) \Longleftrightarrow F\left ( s-a\right )$, then using these into (3) gives $$\begin{align*} \frac{4}{\left ( s+1\right ) ^{3}} & \Longleftrightarrow 4e^{-t}\left ( \frac{t^{2}}{2}\right ) \\ 2\frac{1}{s+1} & \Longleftrightarrow 2e^{-t}\\ 2\frac{1}{\left ( s+1\right ) ^{2}} & \Longleftrightarrow 2e^{-t}t\\ \frac{1}{\left ( s+1\right ) ^{2}} & \Longleftrightarrow e^{-t}t\\ \frac{4}{\left ( s+1\right ) ^{2}} & \Longleftrightarrow 4e^{-t}t \end{align*}$$

Now (3) becomes $$\begin{align*} Y\left ( s\right ) & \Longleftrightarrow 4e^{-t}\left ( \frac{t^{2}}{2}\right ) +2e^{-t}-2e^{-t}t-e^{-t}t+4e^{-t}t\\ & =e^{-t}\left ( 2t^{2}+2-2t-t+4t\right ) \\ & =e^{-t}\left ( 2t^{2}+t+2\right ) \end{align*}$$

2.10.12 Section 6.3 problem 25

Suppose that $F\left ( s\right ) =\mathcal{L}\left \{ f\left ( t\right ) \right \}$ exists for $s>a\geq 0.$

1.

Show that if $c$ is positive constant then $\mathcal{L}\left \{ f\left ( ct\right ) \right \} =\frac{1}{c}F\left ( \frac{s}{c}\right )$ for $s>ca$

2.

Show that if $k$ is positive constant then $\mathcal{L}^{-1}\left \{ F\left ( ks\right ) \right \} =\frac{1}{k}f\left ( \frac{t}{k}\right )$

3.

Show that if $a,b$ are constants with $a>0$ then $\mathcal{L}^{-1}\left \{ F\left ( as+b\right ) \right \} =\frac{1}{a}e^{\frac{-bt}{a}}f\left ( \frac{t}{a}\right )$

Solution

2.10.12.1 Part (a)

From definition, $$\mathcal{L}\left \{ f\left ( ct\right ) \right \} =\int _{0}^{\infty }f\left ( ct\right ) e^{-st}dt$$ Let $ct=\tau$, then when $t=0,\tau =0$ and when $t=\infty ,\tau =\infty$, and $c=\frac{d\tau }{dt}.$ Hence the above becomes $$\begin{align*} \mathcal{L}\left \{ f\left ( ct\right ) \right \} & =\int _{0}^{\infty }f\left ( \tau \right ) e^{-s\left ( \frac{\tau }{c}\right ) }\frac{d\tau }{c}\\ & =\frac{1}{c}\int _{0}^{\infty }f\left ( \tau \right ) e^{-\tau \left ( \frac{s}{c}\right ) }d\tau \end{align*}$$

We see from above that $\mathcal{L}\left \{ f\left ( ct\right ) \right \}$ is $\frac{1}{c}F\left ( \frac{s}{c}\right ) .$Now we look at the conditions which makes the above integral converges. Let $$\left \vert f\left ( \tau \right ) e^{-\tau \left ( \frac{s}{c}\right ) }\right \vert \leq k\left \vert e^{at}e^{-\tau \left ( \frac{s}{c}\right ) }\right \vert$$ Where $k$ is some constant. Then $$\begin{align*} \int _{0}^{\infty }f\left ( t\right ) e^{-t\left ( \frac{s}{c}\right ) }dt & \leq k\int _{0}^{\infty }e^{at}e^{-t\left ( \frac{s}{c}\right ) }dt\\ & =k\int _{0}^{\infty }e^{-t\left ( \frac{s}{c}-a\right ) }dt \end{align*}$$

But $\int _{0}^{\infty }e^{-t\left ( \frac{s}{c}-a\right ) }d\tau$ converges if $\frac{s}{c}-a>0$ or $$s>ca$$ Hence this is the condition for $\int _{0}^{\infty }f\left ( t\right ) e^{-t\left ( \frac{s}{c}\right ) }dt$ to converge. Which is what we required to show.

2.10.12.2 Part (b)

From definition $$\begin{align*} \mathcal{L}\left \{ \frac{1}{k}f\left ( \frac{t}{k}\right ) \right \} & =\frac{1}{k}\mathcal{L}\left \{ f\left ( \frac{t}{k}\right ) \right \} \\ & =\frac{1}{k}\int _{0}^{\infty }f\left ( \frac{t}{k}\right ) e^{-st}dt \end{align*}$$

Let $\frac{t}{k}=\tau$. When $t=0,\tau =0$ and when $t=\infty ,\tau =\infty$. $\frac{dt}{d\tau }=k$, hence the above becomes $$\begin{align*} \mathcal{L}\left \{ \frac{1}{k}f\left ( \frac{t}{k}\right ) \right \} & =\frac{1}{k}\int _{0}^{\infty }f\left ( \tau \right ) e^{-s\left ( k\tau \right ) }\left ( kd\tau \right ) \\ & =\int _{0}^{\infty }f\left ( \tau \right ) e^{-\tau \left ( sk\right ) }d\tau \end{align*}$$

We see from above that $\mathcal{L}\left \{ \frac{1}{k}f\left ( \frac{t}{k}\right ) \right \}$ is $F\left ( sk\right )$. In other words, $\mathcal{L}^{-1}\left \{ F\left ( ks\right ) \right \} =\frac{1}{k}f\left ( \frac{t}{k}\right )$.

2.10.12.3 Part (c)

From definition $$\begin{align*} \mathcal{L}\left \{ \frac{1}{a}e^{\frac{-bt}{a}}f\left ( \frac{t}{a}\right ) \right \} & =\frac{1}{a}\mathcal{L}\left \{ e^{\frac{-bt}{a}}f\left ( \frac{t}{a}\right ) \right \} \\ & =\frac{1}{a}\int _{0}^{\infty }e^{\frac{-bt}{a}}f\left ( \frac{t}{a}\right ) e^{-st}dt \end{align*}$$

Let $\frac{t}{a}=\tau$, at $t=0,\tau =0$ and at $t=\infty ,\tau =\infty$. And $\frac{dt}{d\tau }=a$, hence the above becomes $$\begin{align*} \mathcal{L}\left \{ \frac{1}{a}e^{\frac{-bt}{a}}f\left ( \frac{t}{a}\right ) \right \} & =\frac{1}{a}\int _{0}^{\infty }e^{\frac{-b\left ( a\tau \right ) }{a}}f\left ( \tau \right ) e^{-s\left ( a\tau \right ) }\left ( ad\tau \right ) \\ & =\int _{0}^{\infty }e^{-b\tau }f\left ( \tau \right ) e^{-\tau \left ( sa\right ) }d\tau \\ & =\int _{0}^{\infty }f\left ( \tau \right ) e^{-\tau \left ( sa+b\right ) }d\tau \end{align*}$$

We see from the above, that $\mathcal{L}\left \{ \frac{1}{a}e^{\frac{-bt}{a}}f\left ( \frac{t}{a}\right ) \right \} =F\left ( sa+b\right )$. Now we look at the conditions which makes the above integral converges. Let $$\left \vert f\left ( \tau \right ) e^{-t\left ( sa+b\right ) }\right \vert \leq k\left \vert e^{at}e^{-t\left ( sa+b\right ) }\right \vert$$ Where $k$ is some constant. Then $$\begin{align*} \int _{0}^{\infty }f\left ( t\right ) e^{-t\left ( sa+b\right ) }dt & \leq k\int _{0}^{\infty }e^{at}e^{-t\left ( sa+b\right ) }dt\\ & =k\int _{0}^{\infty }e^{-t\left ( sa+b-a\right ) }dt \end{align*}$$

But $\int _{0}^{\infty }e^{-t\left ( sa+b-a\right ) }dt$ converges if $sa+b-a>0$ or $sa>a-b$ or $s>1-\frac{b}{a}$

2.10.13 Section 6.3 problem 26

Find inverse Laplace transform of $F\left ( s\right ) =\frac{2^{n+1}n!}{s^{n+1}}$

Solution

We know from tables that $$\frac{n!}{s^{n+1}}\Longleftrightarrow t^{n}$$ Hence $$\begin{align*} 2^{n+1}\frac{n!}{s^{n+1}} & \Longleftrightarrow 2^{n+1}t^{n}\\ & =2\left ( 2t\right ) ^{n} \end{align*}$$

2.10.14 Section 6.3 problem 27

Find inverse Laplace transform of $F\left ( s\right ) =\frac{2s+1}{4s^{2}+4s+5}$

Solution $$F\left ( s\right ) =\frac{2s}{4s^{2}+4s+5}+\frac{1}{4s^{2}+4s+5}$$ But $4s^{2}+4s+5=4\left ( s+\frac{1}{2}\right ) ^{2}+4$, hence $$\begin{align} F\left ( s\right ) & =\frac{2s}{4\left ( s+\frac{1}{2}\right ) ^{2}+4}+\frac{1}{4\left ( s+\frac{1}{2}\right ) ^{2}+4}\nonumber \\ & =\frac{s}{2\left ( s+\frac{1}{2}\right ) ^{2}+2}+\frac{1}{4}\frac{1}{\left ( s+\frac{1}{2}\right ) ^{2}+1}\nonumber \\ & =\frac{1}{2}\frac{s}{\left ( s+\frac{1}{2}\right ) ^{2}+1}+\frac{1}{4}\frac{1}{\left ( s+\frac{1}{2}\right ) ^{2}+1}\nonumber \\ & =\frac{1}{2}\frac{s+\frac{1}{2}-\frac{1}{2}}{\left ( s+\frac{1}{2}\right ) ^{2}+1}+\frac{1}{4}\frac{1}{\left ( s+\frac{1}{2}\right ) ^{2}+1}\nonumber \\ & =\frac{1}{2}\frac{s+\frac{1}{2}}{\left ( s+\frac{1}{2}\right ) ^{2}+1}-\frac{1}{4}\frac{1}{\left ( s+\frac{1}{2}\right ) ^{2}+1}+\frac{1}{4}\frac{1}{\left ( s+\frac{1}{2}\right ) ^{2}+1}\nonumber \\ & =\frac{1}{2}\frac{s+\frac{1}{2}}{\left ( s+\frac{1}{2}\right ) ^{2}+1} \tag{1} \end{align}$$

Now we ready to do the inversion. Using $e^{-at}f\left ( t\right ) \Longleftrightarrow F\left ( s+a\right )$ and using $\sin \left ( at\right ) \Longleftrightarrow \frac{a}{s^{2}+a^{2}}$, and $\cos \left ( at\right ) \Longleftrightarrow \frac{s}{s^{2}+a^{2}}$then $$\frac{1}{2}\frac{s+\frac{1}{2}}{\left ( s+\frac{1}{2}\right ) ^{2}+1}\Longleftrightarrow \frac{1}{2}e^{-\frac{1}{2}t}\cos \left ( t\right )$$ Hence $$f\left ( t\right ) =\frac{1}{2}e^{-\frac{1}{2}t}\cos \left ( t\right )$$

2.10.15 Section 6.3 problem 28

Find inverse Laplace transform of $F\left ( s\right ) =\frac{1}{9s^{2}-12s+3}$

Solution $$\frac{1}{9s^{2}-12s+3}=\frac{1}{9}\frac{1}{s^{2}-\frac{4}{3}s+\frac{1}{3}}=\frac{1}{9}\frac{1}{\left ( s-1\right ) \left ( s-\frac{1}{3}\right ) }$$ But $$\begin{align*} \frac{1}{\left ( s-1\right ) \left ( s-\frac{1}{3}\right ) } & =\frac{A}{s-1}+\frac{B}{s-\frac{1}{3}}\\ A & =\left ( \frac{1}{\left ( s-\frac{1}{3}\right ) }\right ) _{s=1}=\frac{3}{2}\\ B & =\left ( \frac{1}{\left ( s-1\right ) }\right ) _{s=\frac{1}{3}}=-\frac{3}{2} \end{align*}$$

Hence $$\begin{equation} \frac{1}{9s^{2}-12s+3}=\frac{1}{9}\left ( \frac{3}{2}\frac{1}{s-1}-\frac{3}{2}\frac{1}{s-\frac{1}{3}}\right ) \tag{1} \end{equation}$$ Using $$e^{at}\Longleftrightarrow \frac{1}{s-a}$$ Then (1) becomes $$\begin{align*} \frac{1}{9s^{2}-12s+3} & \Longleftrightarrow \frac{1}{9}\left ( \frac{3}{2}e^{t}-\frac{3}{2}e^{\frac{1}{3}t}\right ) \\ & =\frac{1}{6}e^{t}-\frac{1}{6}e^{\frac{1}{3}t}\\ & =\frac{1}{6}\left ( e^{t}-e^{\frac{1}{3}t}\right ) \end{align*}$$

2.10.16 Section 6.3 problem 29

Find inverse Laplace transform of $F\left ( s\right ) =\frac{e^{2}e^{-4s}}{2s-1}$

solution

$$F\left ( s\right ) =\frac{e^{2}}{2}\frac{e^{-4s}}{s-\frac{1}{2}}$$ Using $$\begin{equation} u_{c}\left ( t\right ) f\left ( t-c\right ) \Longleftrightarrow e^{-cs}F\left ( s\right ) \tag{1} \end{equation}$$ Since $$\frac{1}{s-\frac{1}{2}}\Longleftrightarrow e^{\frac{1}{2}t}$$ Then using (1) $$e^{-4s}\frac{1}{s-\frac{1}{2}}\Longleftrightarrow u_{4}\left ( t\right ) e^{\frac{1}{2}\left ( t-4\right ) }$$ Hence $$\begin{align*} \frac{e^{2}}{2}\frac{e^{-4s}}{s-\frac{1}{2}} & \Longleftrightarrow \frac{e^{2}}{2}u_{4}\left ( t\right ) e^{\frac{1}{2}\left ( t-4\right ) }\\ & =\frac{1}{2}u_{4}\left ( t\right ) e^{\frac{1}{2}\left ( t-4\right ) +2}\\ & =\frac{1}{2}u_{4}\left ( t\right ) e^{\frac{1}{2}t-2+2}\\ & =\frac{1}{2}u_{4}\left ( t\right ) e^{\frac{t}{2}} \end{align*}$$

Therefore $$f\left ( t\right ) =\frac{1}{2}u_{4}\left ( t\right ) e^{\frac{t}{2}}$$ ps. Book answer is wrong. It gives $$f\left ( t\right ) =\frac{1}{2}u_{4}\left ( \frac{t}{2}\right ) e^{\frac{t}{2}}$$

2.10.17 Section 6.3 problem 30

Find Laplace transform of $f\left ( t\right ) =\left \{ \begin{array} [c]{ccc}1 & & 0\leq t<1\\ 0 & & t\geq 1 \end{array} \right .$

solution

Writing $f\left ( t\right )$ in terms of Heaviside step function gives $$f\left ( t\right ) =u_{0}\left ( t\right ) -u_{1}\left ( t\right )$$ Using $$u_{c}\left ( t\right ) \Longleftrightarrow e^{-cs}\frac{1}{s}$$ Therefore $$\begin{align*} \mathcal{L}\left \{ u_{0}\left ( t\right ) \right \} & =e^{-0s}\frac{1}{s}=\frac{1}{s}\\\mathcal{L}\left \{ u_{1}\left ( t\right ) \right \} & =e^{-s}\frac{1}{s} \end{align*}$$

Hence $$\begin{align*} \mathcal{L}\left \{ u_{0}\left ( t\right ) -u_{1}\left ( t\right ) \right \} & =\frac{1}{s}-e^{-s}\frac{1}{s}\\ & =\frac{1}{s}\left ( 1-e^{-s}\right ) \qquad s>0 \end{align*}$$

2.10.18 Section 6.3 problem 31

Find Laplace transform of $f\left ( t\right ) =\left \{ \begin{array} [c]{ccc}1 & & 0\leq t<1\\ 0 & & 1\leq t<2\\ 1 & & 2\leq t<3\\ 0 & & t\geq 3 \end{array} \right .$

solution

Writing $f\left ( t\right )$ in terms of Heaviside step function gives $$f\left ( t\right ) =u_{0}\left ( t\right ) -u_{1}\left ( t\right ) +u_{2}\left ( t\right ) -u_{3}\left ( t\right )$$ Using $$u_{c}\left ( t\right ) \Longleftrightarrow e^{-cs}\frac{1}{s}$$ But $f\left ( t\right ) =1$ in this case. Hence $F\left ( s\right ) =\frac{1}{s}$. Therefore $$\begin{align*} f\left ( t\right ) & \Longleftrightarrow \frac{1}{s}e^{0s}-\frac{1}{s}e^{-s}+\frac{1}{s}e^{-2s}-\frac{1}{s}e^{-3s}\\ & =\frac{1}{s}\left ( 1-e^{-s}+e^{-2s}-e^{-3s}\right ) \qquad s>0 \end{align*}$$

2.10.19 Section 6.3 problem 32

Find Laplace transform of $f\left ( t\right ) =1+\sum _{k=1}^{2n+1}\left ( -1\right ) ^{k}u_{k}\left ( t\right )$

solution

Using $$u_{c}\left ( t\right ) \Longleftrightarrow e^{-cs}\frac{1}{s}$$ Therefore $$\begin{align*} \mathcal{L}\left \{ 1+\sum _{k=1}^{2n+1}\left ( -1\right ) ^{k}u_{k}\left ( t\right ) \right \} & =\mathcal{L}\left \{ 1\right \} +\mathcal{L}\left \{ \sum _{k=1}^{2n+1}\left ( -1\right ) ^{k}u_{k}\left ( t\right ) \right \} \\ & =\frac{1}{s}+\sum _{k=1}^{2n+1}\left ( -1\right ) ^{k}\frac{1}{s}e^{-ks}\\ & =\sum _{k=0}^{2n+1}\left ( -1\right ) ^{k}\frac{1}{s}e^{-ks}\\ & =\frac{1}{s}\sum _{k=0}^{2n+1}\left ( -e^{-s}\right ) ^{k} \end{align*}$$

Since $\left \vert e^{-s}\right \vert <1$ the sum converges. Using $\sum _{0}^{N}a_{n}=\left ( \frac{1-r^{N+1}}{1-r}\right )$. Where $\left \vert r\right \vert <1$. So the answer is $$\begin{align*} \mathcal{L}\left \{ 1+\sum _{k=1}^{2n+1}\left ( -1\right ) ^{k}u_{k}\left ( t\right ) \right \} & =\frac{1}{s}\left ( \frac{1-\left ( -e^{-s}\right ) ^{2n+2}}{1-\left ( -e^{-s}\right ) }\right ) \\ & =\frac{1}{s}\left ( \frac{1-\left ( -e\right ) ^{-\left ( 2n+2\right ) s}}{1+e^{-s}}\right ) \end{align*}$$

Since $2n+2$ is even then $$\mathcal{L}\left \{ 1+\sum _{k=1}^{2n+1}\left ( -1\right ) ^{k}u_{k}\left ( t\right ) \right \} =\frac{1}{s}\left ( \frac{1+e^{-\left ( 2n+2\right ) s}}{1+e^{-s}}\right ) \qquad s>0$$

2.10.20 Section 6.3 problem 33

Find Laplace transform of $f\left ( t\right ) =1+\sum _{k=1}^{\infty }\left ( -1\right ) ^{k}u_{k}\left ( t\right )$

solution

Using $$u_{c}\left ( t\right ) \Longleftrightarrow e^{-cs}\frac{1}{s}$$ Therefore $$\begin{align*} \mathcal{L}\left \{ 1+\sum _{k=1}^{\infty }\left ( -1\right ) ^{k}u_{k}\left ( t\right ) \right \} & =\mathcal{L}\left \{ 1\right \} +\mathcal{L}\left \{ \sum _{k=1}^{\infty }\left ( -1\right ) ^{k}u_{k}\left ( t\right ) \right \} \\ & =\frac{1}{s}+\sum _{k=1}^{\infty }\left ( -1\right ) ^{k}\frac{1}{s}e^{-ks}\\ & =\frac{1}{s}+\frac{1}{s}\sum _{k=1}^{\infty }\left ( -1\right ) ^{k}e^{-ks}\\ & =\frac{1}{s}+\frac{1}{s}\sum _{k=1}^{\infty }\left ( -e^{-s}\right ) ^{k} \end{align*}$$

But $$\sum _{k=1}^{\infty }r^{k}=\frac{r}{1-r}\qquad \left \vert r\right \vert <1$$ Since $s>0$ then$\left \vert e^{-s}\right \vert <1$. So the answer is $$\begin{align*} \frac{1}{s}+\frac{1}{s}\frac{-e^{-s}}{1-\left ( -e^{-s}\right ) } & =\frac{1}{s}-\frac{1}{s}\frac{e^{-s}}{1+e^{-s}}\\ & =\frac{1+e^{-s}-e^{-s}}{s\left ( 1+e^{-s}\right ) }\\ & =\frac{1}{s\left ( 1+e^{-s}\right ) }\qquad s>0 \end{align*}$$

2.10.21 Section 6.4 problem 21

2.10.21.1 Part (a)

A plot of part (a) is the following

And a plot of part(a) for problem 19 is the following

We see the effect of having a $2$ inside the sum. It extends the step $u_{c}\left ( t\right )$ function to negative side.

2.10.21.2 Part (b)

The easy way to do this, is to solve for each input term separately, and then add all the solutions, since this is a linear ODE. Once we solve for the first 2-3 terms, we will see the pattern to use for the overall solution. Since the input $g\left ( t\right )$ is $u_{0}\left ( t\right ) +\sum _{k=1}^{\infty }\left ( -1\right ) ^{k}u_{k\pi }\left ( t\right )$, we will first first the response to $u_{0}\left ( t\right ) \,$, then for $-u_{\pi }\left ( t\right )$ then for $+u_{2\pi }\left ( t\right )$, and so on, and add them.

When the input is $u_{0}\left ( t\right )$, then its Laplace transform is $\frac{1}{s}$, Hence, taking Laplace transform of the ODE gives (where now $Y\left ( s\right ) =\mathcal{L}\left ( y\left ( t\right ) \right )$) $$\left ( s^{2}Y\left ( s\right ) -sy\left ( 0\right ) +y^{\prime }\left ( 0\right ) \right ) +Y\left ( s\right ) =\frac{1}{s}$$ Applying initial conditions $$s^{2}Y\left ( s\right ) +Y\left ( s\right ) =\frac{1}{s}$$ Solving for $Y_{0}\left ( s\right )$ (called it $Y_{0}\left ( s\right ) \,$ since the input is $u_{0}\left ( t\right )$) $$\begin{align*} Y_{0}\left ( s\right ) & =\frac{1}{s\left ( s^{2}+1\right ) }\\ & =\frac{1}{s}-\frac{s}{s^{2}+1} \end{align*}$$

Hence $$y_{0}\left ( t\right ) =1-\cos t$$ We now do the next input, which is $-u_{\pi }\left ( t\right )$, which has Laplace transform of $-\frac{e^{-\pi s}}{s}$, therefore, following what we did above, we obtain now $$\begin{align*} Y_{\pi }\left ( s\right ) & =\frac{-e^{-\pi s}}{s\left ( s^{2}+1\right ) }\\ & =-e^{-\pi s}\left ( \frac{1}{s}-\frac{s}{s^{2}+1}\right ) \end{align*}$$

The effect of $e^{-\pi s}$ is to cause delay in time. Hence the the inverse Laplace transform of the above is the same as $y_{0}\left ( t\right )$ but with delay $$y_{\pi }\left ( t\right ) =-u_{\pi }\left ( t\right ) \left ( 1-\cos \left ( t-\pi \right ) \right )$$ Similarly, when the input is $+u_{2\pi }\left ( t\right )$, which which has Laplace transform of $\frac{e^{-2\pi s}}{s}$, therefore, following what we did above, we obtain now $$\begin{align*} Y_{\pi }\left ( s\right ) & =\frac{e^{-2\pi s}}{s\left ( s^{2}+1\right ) }\\ & =e^{-2\pi s}\left ( \frac{1}{s}-\frac{s}{s^{2}+1}\right ) \end{align*}$$

The effect of $e^{-2\pi s}$ is to cause delay in time. Hence the the inverse Laplace transform of the above is the same as $y_{0}\left ( t\right )$ but with now with delay of $2\pi$, therefore $$y_{2\pi }\left ( t\right ) =+u_{2\pi }\left ( t\right ) \left ( 1-\cos \left ( t-2\pi \right ) \right )$$ And so on. We see that if we add all the responses, we obtain $$\begin{align*} y\left ( t\right ) & =y_{0}\left ( t\right ) +y_{\pi }\left ( t\right ) +y_{2\pi }\left ( t\right ) +\cdots \\ & =\left ( 1-\cos t\right ) -u_{\pi }\left ( t\right ) \left ( 1-\cos \left ( t-\pi \right ) \right ) +u_{2\pi }\left ( t\right ) \left ( 1-\cos \left ( t-2\pi \right ) \right ) -\cdots \end{align*}$$

Or $$\begin{equation} y\left ( t\right ) =\left ( 1-\cos t\right ) +\sum _{k=1}^{n}\left ( -1\right ) ^{k}u_{k\pi }\left ( t\right ) \left ( 1-\cos \left ( t-k\pi \right ) \right ) \tag{1} \end{equation}$$

2.10.21.3 Part (c)

This is a plot of (1) for $n=15$

We see the solution growing rapidly, they settling down after about $t=50$ to sinusoidal wave at amplitude of about $\pm 15$.  This shows the system reached steady state at around $t=50$.

To compare it with problem 19 solution, I used the solution for 19 given in the book, and plotted both solution on top of each others. Also for up to $t=60$. Here is the result

We see that problem 19 output follows the same pattern (since same frequency is used), but with double the amplitude. This is due to the $2$ factor used in problem 19 compared to this problem.

2.10.21.4 Part(d)

At first, I tried it with $n=50,150,250,350,450,550.$ I can not see any noticeable change in the plot. Here is the result.

Even at $n=2000$ there was no change to be noticed.

This shows additional input in the form of shifted unit steps, do not change the steady state solution.