Problem Find the inverse transform of the function \(F\left ( p\right ) =\frac {3p+2}{3p^{2}+5p-2}\)
Solution
Need to simplify the above expression to some expressions which are shown in the table on page 636.
\begin {align} F(p) & =\frac {3p+2}{3p^{2}+5p-2}\nonumber \\ & =\frac {2}{\left (3p-1\right ) \left (p+2\right ) }+\frac {3p}{\left ( 3p-1\right ) \left (p+2\right ) } \tag {1} \end {align}
Expanding in partial fractions. For the first term in (1):
\begin {align*} \frac {1}{\left (3p-1\right ) \left (p+2\right ) } & =\frac {A}{\left ( 3p-1\right ) }+\frac {B}{\left (p+2\right ) }=\frac {A\left (p+2\right ) +B\left (3p-1\right ) }{\left (3p-1\right ) \left (p+2\right ) }\\ A\left (p+2\right ) +B\left (3p-1\right ) & =1\\ Ap+2A+3Bp-B & =1 \end {align*}
Hence \(2A-B=1\) and \(\left (A+3B\right ) =0\) which gives \(A=\frac {1}{2}+\frac {B}{2}\). Therefore \(\frac {1}{2}+\frac {B}{2}+3B=0\) \(\ \) or \(\ \ \frac {1+B+6B}{2}\ =0\) or \(1+7B=0\) or \(B=\frac {-1}{7}\). Hence \(A=\frac {1}{2}+\frac {\frac {-1}{7}}{2}=\frac {1}{2}-\frac {1}{14}=\frac {\ 7-1}{14}=\frac {\ 6}{14}\).
Now the first term in (1) can be written as \(2\left (\frac {A}{\left ( 3p-1\right ) }+\frac {B}{\left (p+2\right ) }\right ) \) or \(2\left ( \frac {\frac {\ 6}{14}}{\left (3p-1\right ) }+\frac {\frac {-1}{7}}{\left ( p+2\right ) }\right ) \) or
\begin {equation} \frac {\ 6}{7\left (3p-1\right ) }-\frac {2}{7\left (p+2\right ) } \tag {2} \end {equation}
Doing partial fraction on the second term in (1) which is \(\frac {3p}{\left ( 3p-1\right ) \left (p+2\right ) }\) gives
\begin {align*} \frac {p}{\left (3p-1\right ) \left (p+2\right ) } & =\frac {A}{\left ( 3p-1\right ) }+\frac {B}{\left (p+2\right ) }=\frac {A\left (p+2\right ) +B\left (3p-1\right ) }{\left (3p-1\right ) \left (p+2\right ) }\\ A\left (p+2\right ) +B\left (3p-1\right ) & =p\\ Ap+2A+3Bp-B & =p \end {align*}
Hence \(2A-B=0\) and \(\left (A+3B\right ) =1\), therefore \(A=\frac {B}{2}\). Hence \(\left (\frac {B}{2}+3B\right ) =1\) or \(\frac {B+6B}{2}=1\) or \(7B=2\) or \(B=\frac {2}{7}\). Therefore \(A=\frac {B}{2}=\frac {2}{14}\). Hence \(\frac {3p}{\left (3p-1\right ) \left (p+2\right ) }=3\left (\frac {A}{\left ( 3p-1\right ) }+\frac {B}{\left (p+2\right ) }\right ) =3\left (\frac {\frac {2}{14}}{\left (3p-1\right ) }+\frac {\frac {2}{7}}{\left (p+2\right ) }\right ) \) or
\begin {equation} \frac {3}{7\left (3p-1\right ) }+\frac {6}{7\left (p+2\right ) } \tag {3} \end {equation}
Combining (2) and (3) gives
\begin {align*} F(p) & =\frac {\ 6}{7\left (3p-1\right ) }-\frac {2}{7\left (p+2\right ) }+\frac {3}{7\left (3p-1\right ) }+\frac {6}{7\left (p+2\right ) }\\ & =\frac {6}{7}\frac {\ 1}{\left (3p-1\right ) }-\frac {2}{7}\frac {1}{\left ( p+2\right ) }+\frac {2}{7}\frac {1}{\left (3p-1\right ) }+\frac {6}{7}\frac {1}{\left (p+2\right ) }\\ & =\frac {4}{7}\frac {1}{\left (p+2\right ) }+\frac {9}{7}\frac {\ 1}{\left ( 3p-1\right ) }\\ & =\frac {4}{7}\frac {1}{\left (p+2\right ) }+\frac {9}{21}\frac {\ 1}{\left ( p-\frac {1}{3}\right ) } \end {align*}
Now we can use the table to find the inverse transform. Use property L2, which says \[\mathcal {L}\left (e^{-at}\right ) =\frac {1}{p+a}\] Hence , setting \(a=2,\) gives \(\mathcal {L}\left (e^{-2t}\right ) =\frac {1}{p+2}\) and setting \(a=-\frac {1}{3}\) gives \(\mathcal {L}\left (e^{-\frac {1}{3}t}\right ) ^{\frac {1}{3}t}=\frac {1}{p-\frac {1}{3}}\). Hence \(f\relax (t) =\frac {4}{7}e^{-2t}+\frac {9}{21}e^{\frac {1}{3}t}\) and the inverse Laplace transform is \[ \ f\relax (t) =\frac {4}{7}e^{-2t}+\frac {3}{7}e^{\frac {1}{3}t}\]
Problem Use L32 and L11 to obtain \(\mathcal {L}\left (t^{2}\ \sin at\right ) \)
Solution
\begin {align} \mathcal {L}\left (t^{n}f\relax (t) \right ) & =\left (-1\right ) ^{n}\frac {d^{n}F\relax (p) }{dp^{n}}\tag {L32}\\\mathcal {L}\left (t\ \sin at\right ) & =\ \frac {2ap}{\left (p^{2}+a^{2}\right ) ^{2}} \tag {L11} \end {align}
we set \(f\relax (t) =t\ \sin at\) then we can write using L32
\begin {equation} \mathcal {L}\left (t\ f\relax (t) \right ) =\left (-1\right ) \frac {d\mathcal {L}\left (f\relax (t) \right ) }{dp} \tag {1} \end {equation}
But \(\mathcal {L}\left (f\relax (t) \right ) =\mathcal {L}\left (t\ \sin at\right ) \) \(=\ \frac {2ap}{\left (p^{2}+a^{2}\right ) ^{2}}\ \)from table L11 (1) becomes
\begin {align*} \mathcal {L}\left (t\ f\relax (t) \right ) & =-\frac {d}{dp}\left (\frac {2ap}{\left (p^{2}+a^{2}\right ) ^{2}}\right ) \\\mathcal {L}\left (t\times t\ \sin at\right ) & =\ -\left (p\ \frac {-2\times 2a}{\left (p^{2}+a^{2}\right ) ^{3}}\times 2p+\frac {2a}{\left (p^{2}+a^{2}\right ) ^{2}}\times 1\right ) \\\mathcal {L}\left (t^{2}\ \sin at\right ) & =\ \frac {8ap^{2}}{\left (p^{2}+a^{2}\right ) ^{3}}-\frac {2a}{\left (p^{2}+a^{2}\right ) ^{2}}\\ & =\ \frac {8ap^{2}-2a\left (p^{2}+a^{2}\right ) }{\left (p^{2}+a^{2}\right ) ^{3}}\ \\ & =\frac {a\left (8p^{2}-2p^{2}-2a^{2}\right ) }{\left (p^{2}+a^{2}\right ) ^{3}}\\ & =\frac {a\left (6p^{2}-2a^{2}\right ) }{\left (p^{2}+a^{2}\right ) ^{3}} \end {align*}
Or
\[\mathcal {L}\left (t^{2}\ \sin at\right ) =\frac {6ap^{2}-2a^{3}}{\left (p^{2}+a^{2}\right ) ^{3}}\]
Problem Use L31 to derive L21
Solution
\begin {align} \mathcal {L}\left (\frac {f\relax (t) }{t}\right ) & =\ \int _{u=p}^{\infty }F\relax (u) du\tag {L31}\\\mathcal {L}\left (\ \frac {e^{-at}-e^{-bt}}{t}\right ) & =\ \ln \frac {p+b}{p+a} \tag {L21} \end {align}
we set \(f\relax (t) =e^{-at}-e^{-bt}\) then we can write using L31
\begin {align*} \mathcal {L}\left (\frac {f\relax (t) }{t}\right ) & =\int _{u=p}^{\infty }F\left ( u\right ) du\\ & =\int _{u=p}^{\infty }\mathcal {L}\left [ f\relax (t) \right ] \ \ du \end {align*}
but \(\mathcal {L}\left [ f\relax (t) \right ] =\mathcal {L}\left (e^{-at}-e^{-bt}\right ) =\mathcal {L}\left (e^{-at}\right ) -\mathcal {L}\left (e^{-bt}\right ) =\frac {1}{p+a}-\frac {1}{p+b}\) By using L2. But since we are using \(u\) in place of \(p\) in integral, we need to call \(p=u\). Hence
\begin {align*} \mathcal {L}\left (\frac {f\relax (t) }{t}\right ) & =\int _{u=p}^{\infty }\left ( \frac {1}{u+a}-\frac {1}{u+b}\right ) \ \ du\\ & =\int _{u=p}^{\infty }\frac {1}{u+a}\ du-\int _{u=p}^{\infty }\frac {1}{u+b}\ \ du\\ & =\left [ \ln \left (u+a\right ) \right ] _{p}^{\infty }-\left [ \ln \left ( u+b\right ) \right ] _{p}^{\infty }\\ & =\left [ \ln \left (\infty +a\right ) -\ln \left (p+a\right ) \right ] -\left [ \ln \left (\infty +b\right ) -\ln \left (p+b\right ) \right ] \\ & =\ln \left (\infty \right ) -\ln \left (p+a\right ) -\ln \left ( \infty \right ) +\ln \left (p+b\right ) \\ & =\ln \left (p+b\right ) -\ln \left (p+a\right ) \end {align*}
but \(\ln A-\ln B=\ln \frac {A}{B}\), therefore
\begin {align*} \mathcal {L}\left (\frac {f\relax (t) }{t}\right ) & =\ \ln \left (p+b\right ) -\ln \left (p+a\right ) \\\mathcal {L}\left (\frac {e^{-at}-e^{-bt}}{t}\right ) & =\ln \left (\frac {p+b}{p+a}\right ) \end {align*}
Problem Use relation L2 to find L7 and L8 in laplace table.
Solution
\begin {equation} \mathcal {L}\left (e^{-at}\right ) =\frac {1}{p+a} \tag {L2} \end {equation}
for \(\Re \left (p+a\right ) >0\)
\begin {equation} \mathcal {L}\frac {e^{-at}-e^{-bt}}{b-a}=\frac {1}{\left (p+a\right ) \left (p+b\right ) } \tag {L7} \end {equation}
\begin {equation} \mathcal {L}\frac {ae^{-at}-be^{-bt}}{a-b}=\frac {p}{\left (p+a\right ) \left (p+b\right ) } \tag {L8} \end {equation}
For \(\Re \left (p+a\right ) >0\) and \(\Re \left (p+b\right ) >0\). Where \(\mathcal {L}\)\(f(t)\) is the laplace transform of \(f\relax (t) \) defined as \(\mathcal {L}\)\(f\relax (t) =F\relax (p) =\int _{0}^{\infty }e^{-pt}f\left ( t\right ) dt.\)
From the linearity property of the \(\mathcal {L}\) operator, expand the LHS of L7, we get
\[\mathcal {L}\frac {e^{-at}-e^{-bt}}{b-a}=\frac {1}{b-a}\mathcal {L}\left (e^{-at}-e^{-bt}\right ) =\frac {1}{b-a}\left ( \mathcal {L}\left (e^{-at}\right ) -\mathcal {L}\left (e^{-bt}\right ) \right ) \]
Now applying L2 gives for \(\Re \left (p+a\right ) >0\) and \(\Re \left ( p+b\right ) >0\)
\begin {align*} \mathcal {L}\frac {e^{-at}-e^{-bt}}{b-a} & =\frac {1}{b-a}\left (\frac {1}{p+a}-\frac {1}{p+b}\right ) \\ & =\frac {1}{b-a}\left (\ \frac {p+b-\left (p+a\right ) }{\left (p+a\right ) \left (p+b\right ) }\right ) \\ & =\frac {1}{b-a}\left (\ \frac {b-a}{\left (p+a\right ) \left (p+b\right ) }\right ) \\ & =\frac {1}{\left (p+a\right ) \left (p+b\right ) } \end {align*}
For Which is L7 as required to show. Similarly for L8, expand the LHS of L8 we get for \(\Re \left (p+a\right ) >0\) and \(\Re \left (p+b\right ) >0\)
\begin {align*} \mathcal {L}\frac {ae^{-at}-be^{-bt}}{a-b} & =\frac {1}{a-b}\mathcal {L}\left (ae^{-at}-be^{-bt}\right ) \\ & =\frac {1}{a-b}\left ( \mathcal {L}\left (e^{-at}\right ) -\mathcal {L}\left (e^{-bt}\right ) \right ) \\ & =\frac {1}{a-b}\left (a\mathcal {L}\left (e^{-at}\right ) -b\mathcal {L}\left (e^{-bt}\right ) \right ) \\ & =\frac {1}{a-b}\left (a\frac {1}{p+a}-b\frac {1}{p+b}\right ) \\ & =\frac {1}{a-b}\left (\frac {a\left (p+b\right ) -b\left (p+a\right ) }{\left (p+a\right ) \left (p+b\right ) }\right ) \\ & =\frac {1}{a-b}\left (\frac {ap+ab-bp-ba}{\left (p+a\right ) \left ( p+b\right ) }\right ) \\ & =\frac {1}{a-b}\left (\frac {ap-bp}{\left (p+a\right ) \left (p+b\right ) }\right ) \\ & =\frac {1}{a-b}\left (\frac {p\left (a-b\right ) }{\left (p+a\right ) \left (p+b\right ) }\right ) \\ & =\frac {p}{\left (p+a\right ) \left (p+b\right ) } \end {align*}
Which is L8.
Problem Use L29 and L11 to obtain \(\mathcal {L}\left (te^{-at}\ \sin bt\right ) \)
Solution
\begin {align} \mathcal {L}\left (e^{-at}\ f\relax (t) \right ) & =\ F\left (p+a\right ) \tag {L29}\\\mathcal {L}\left (t\ \sin at\right ) & =\ \ \frac {2ap\ }{\left (p^{2}+a^{2}\right ) ^{2}} \tag {L11} \end {align}
Then from L11 we get
\[\mathcal {L}\left (t\ \sin bt\right ) =\ \ \frac {2bp\ }{\left (p^{2}+b^{2}\right ) ^{2}}\]
Now, let \(p=\left (p+a\right ) \) then from L29, the above becomes
\[\mathcal {L}\left (e^{-at}\ t\ \sin bt\right ) =\ \ \frac {2b\left (p+a\right ) \ }{\left (\left (p+a\right ) ^{2}+b^{2}\right ) ^{2}}\]
Problem similar to problem 2.21, Use L29 and L12 to obtain \(\mathcal {L}\left (te^{-at}\ \cos bt\right ) \)
Solution
\begin {align} \mathcal {L}\left (e^{-at}\ f\relax (t) \right ) & =\ F\left (p+a\right ) \tag {L29}\\\mathcal {L}\left (t\ \cos at\right ) & =\ \ \frac {p^{2}-a^{2}\ }{\left (p^{2}+a^{2}\right ) ^{2}} \tag {L12} \end {align}
then from L12 we get
\[\mathcal {L}\left (t\ \cos bt\right ) =\ \ \frac {p^{2}-b^{2}\ }{\left (p^{2}+b^{2}\right ) ^{2}}\]
Now, let \(p=\left (p+a\right ) \) then from L29, the above becomes
\[\mathcal {L}\left (e^{-at}\ t\ \cos bt\right ) =\ \ \frac {\left (p+a\right ) ^{2}-b^{2}\ }{\left (\left (p+a\right ) ^{2}+b^{2}\right ) ^{2}}\]
Problem use result obtained in problem 2.21 and 2.22 to find inverse transform for \(\frac {p^{2}+2p-1}{\left (p^{2}+4p+5\right ) ^{2}}\)
Solution
Recall, from 2.21 we showed that \(\mathcal {L}\left (te^{-at}\ \sin bt\right ) =\frac {2b\left (p+a\right ) \ }{\left ( \left (p+a\right ) ^{2}+b^{2}\right ) ^{2}}\) and from 2.22 \(\mathcal {L}\left (te^{-at}\ \cos bt\right ) =\frac {\left (p+a\right ) ^{2}-b^{2}\ }{\left (\left (p+a\right ) ^{2}+b^{2}\right ) ^{2}}\) Hence\begin {align*} \mathcal {L}\left (te^{-at}\ \cos bt\right ) -\mathcal {L}\left (te^{-at}\ \sin bt\right ) & =\frac {\left (p+a\right ) ^{2}-b^{2}\ }{\left (\left (p+a\right ) ^{2}+b^{2}\right ) ^{2}}-\frac {2b\left ( p+a\right ) \ }{\left (\left (p+a\right ) ^{2}+b^{2}\right ) ^{2}}\\ & =\frac {\left (p+a\right ) ^{2}+b^{2}\ -2b\left (p+a\right ) }{\left ( \left (p+a\right ) ^{2}+b^{2}\right ) ^{2}}\ \end {align*}
Now Let \(a=2\) and let \(b=1\) we get
\begin {align*} \mathcal {L}\left (te^{-2t}\ \cos t\right ) -\mathcal {L}\left (te^{-2t}\ \sin t\right ) & =\frac {\left (p+2\right ) ^{2}-1^{2}-2\left (p+2\right ) \ }{\left (\left (p+2\right ) ^{2}+1^{2}\right ) ^{2}}\\\mathcal {L}\left (te^{-2t}\ \cos t-te^{-2t}\ \sin t\right ) \ & =\ \frac {p^{2}+4+4p-1-2p-4\ }{\left (\left (p+2\right ) ^{2}+1^{2}\right ) ^{2}}\\ & =\frac {\ p^{2}+2p-1\ \ }{\left (p^{2}+4p+5\right ) ^{2}} \end {align*}
Hence inverse transform of \(\frac {\ p^{2}+2p-1\ \ }{\left (p^{2}+2p+5\right ) ^{2}}\) is \(te^{-2t}\ \cos t-te^{-2t}\ \sin t\) or \(\ te^{-2t}\ \left (\cos t-\sin t\right ) \)
Problem Using either relation L2 or L3 and L4, verify L9 and L10 in laplace table.
Solution
\begin {equation} \mathcal {L}\left (e^{-at}\right ) =\frac {1}{p+a}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \operatorname {Re}\left ( p+a\right ) >0 \tag {L2} \end {equation}
\begin {equation} \mathcal {L}\sin at=\frac {a}{p^{2}+a^{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \operatorname {Re}\relax (p) >\left \vert \operatorname {Im}a\right \vert \ \tag {L3} \end {equation}
\begin {equation} \mathcal {L}\cos at=\frac {p}{p^{2}+a^{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \operatorname {Re}\relax (p) >\left \vert \operatorname {Im}a\right \vert \ \tag {L4} \end {equation}
\begin {equation} \mathcal {L}\sinh at=\frac {a}{p^{2}-a^{2}\ }\ \ \ \ \ \ \ \ \ \ \ \ \operatorname {Re}\relax (p) >\left \vert \ \operatorname {Re}\relax (a) \right \vert \ \ \tag {L9} \end {equation}
\begin {equation} \mathcal {L}\cosh at=\frac {p}{p^{2}-a^{2}\ }\ \ \ \ \ \ \ \ \ \ \ \ \operatorname {Re}\relax (p) >\left \vert \ \operatorname {Re}\relax (a) \right \vert \ \ \tag {L10} \end {equation}
Where \(\mathcal {L}\left (f\relax (t) \right ) \) is the laplace transform of \(f\left ( t\right ) \) defined as \(\mathcal {L}\left (f\relax (t) \right ) =Y\relax (s) =\int _{0}^{\infty }e^{-pt}f\relax (t) dt\). To derive L9, use the relation that
\[ i\sinh \relax (x) =\sin \left (ix\right ) \]
Hence, using L3, we get
\begin {align*} \mathcal {L}\left (i\sinh at\right ) & =\mathcal {L}\left (\sin iat\ \right ) \\ i\mathcal {L}\left (\sinh at\right ) & =\frac {ia}{p^{2}+\left (ia\right ) ^{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \operatorname {Re}\relax (p) >\left \vert \operatorname {Im}a\right \vert \\\mathcal {L}\left (\sinh at\right ) & =\ \frac {a}{p^{2}-a^{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \operatorname {Re}\relax (p) >\left \vert \ \operatorname {Re}\relax (a) \right \vert \ \ \ \ \ \ \ \ \ \ \end {align*}
which is L9. To find L10, use the relation
\[ \cosh \relax (x) =\cos \left (ix\right ) \]
And using L4, we get
\begin {align*} \mathcal {L}\left (\cosh \left (at\right ) \right ) & =\mathcal {L}\left (\cos \left (iat\right ) \ \right ) \\\mathcal {L}\left (\cosh at\right ) & =\frac {p}{p^{2}+\left (ia\right ) ^{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \operatorname {Re}\relax (p) >\left \vert \operatorname {Im}a\right \vert \\\mathcal {L}\left (\sinh at\right ) & =\ \frac {p}{p^{2}-a^{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \operatorname {Re}\relax (p) >\left \vert \ \operatorname {Re}\relax (a) \right \vert \ \ \ \ \ \ \end {align*}
Which is L10.
Problem by differentiating the appropriate formulas w.r.t. ’a’, verify L12
Solution
L12 is
\begin {equation} \mathcal {L}\left (t\ \cos t\right ) =\frac {p^{2}-a^{2}}{\left (p^{2}+a^{2}\right ) ^{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \operatorname {Re}\left ( p\right ) >\left \vert \operatorname {Im}a\right \vert \ \tag {L12} \end {equation}
Where \(\mathcal {L}\left (f\relax (t) \right ) \) is the laplace transform of \(f\left ( t\right ) \) defined as\(\mathcal {L}\left (f\relax (t) \right ) =F\relax (p) =\int _{0}^{\infty }e^{-pt}f\relax (t) dt\). To derive this, I start with L3, which says
\[\mathcal {L}\left (\sin at\right ) =\frac {a}{p^{2}+a^{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \operatorname {Re}\relax (p) >\left \vert \operatorname {Im}a\right \vert \]
The above can be rewritten in the full definition of the transform to make it easier to see
\[ \int _{0}^{\infty }e^{-pt}\ \sin at\ \ dt=\frac {a}{p^{2}+a^{2}}\]
Taking derivative of both sides w.r.t. \(a\) gives
\begin {align*} \frac {d}{da}\left [ \int _{0}^{\infty }e^{-pt}\ \sin at\ \ dt\right ] & =\frac {d}{da}\left [ \frac {a}{p^{2}+a^{2}}\right ] \\ \int _{0}^{\infty }\frac {d}{da}\left [ e^{-pt}\ \sin at\right ] \ \ dt & =a\left (\frac {-2a}{\left (p^{2}+a^{2}\right ) ^{2}}\right ) +\frac {1}{p^{2}+a^{2}}\times 1\\ \int _{0}^{\infty }e^{-pt}\ t\cos at\ \ dt & =\frac {-2a^{2}}{\left ( p^{2}+a^{2}\right ) ^{2}}+\frac {1}{p^{2}+a^{2}}\\\mathcal {L}\left (t\ \cos at\right ) & =\frac {-2a^{2}+\left (p^{2}+a^{2}\right ) }{\left (p^{2}+a^{2}\right ) ^{2}}\ \\ \ & =\frac {p^{2}-a^{2}}{\left (p^{2}+a^{2}\right ) ^{2}}\ \end {align*}
Which is L12
Problem by integrating the appropriate formulas w.r.t. ’a’, verify L19
Solution L19 is
\begin {equation} \mathcal {L}\left (\frac {\sin at}{t}\right ) =\arctan \frac {a}{p}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \operatorname {Re}\relax (p) >\left \vert \operatorname {Im}a\right \vert \ \tag {L19} \end {equation}
Where \(\mathcal {L}\left (f\relax (t) \right ) \) is the laplace transform of \(f\left ( t\right ) \) defined as \(\mathcal {L}\left (f\relax (t) \right ) =F\relax (p) =\int _{0}^{\infty }e^{-pt}f\relax (t) dt\). To derive this, we start with L4, which says
\[\mathcal {L}\left (\cos at\right ) =\frac {p}{p^{2}+a^{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \operatorname {Re}\relax (p) >\left \vert \operatorname {Im}a\right \vert \]
The above can be rewritten in the full definition of the transform to make it easier to see
\[ \int _{0}^{\infty }e^{-pt}\ \cos at\ \ dt=\frac {p}{p^{2}+a^{2}}\]
Integrating both sides w.r.t. \(a\) gives
\begin {align*} \int \left (\int _{t=0}^{t=\infty }e^{-pt}\ \cos at\ \ dt\right ) da & =\int \frac {p}{p^{2}+a^{2}}da\\ \int _{t=0}^{t=\infty }\left (\int e^{-pt}\ \cos at\ da\right ) \ dt & =\arctan \frac {a}{p}+K\\ \int _{t=0}^{t=\infty }e^{-pt}\ \left (\int \cos at\ da\right ) \ dt & =\arctan \frac {a}{p}+K\ \end {align*}
Now, since \(\int \cos at\ da=\frac {\sin at}{t}+k\) and choose zero for values of the K’s, the constants of integration gives
\[ \int _{0}^{\infty }e^{-pt}\ \left (\ \frac {\sin at}{t}\right ) \ dt=\arctan \frac {a}{p}\]
Hence
\[\mathcal {L}\left (\frac {\sin at}{t}\right ) =\arctan \frac {a}{p}\ \ \]
Which is L19
Problem Find the inverse transform of the function \(F\left ( p\right ) =\frac {5-2p}{p^{2}+p-2}\)
Solution
Need to simplify the above expression to some expressions which are shown in the table on page 636.
\begin {align} F(p) & =\frac {5-2p}{p^{2}+p-2}\nonumber \\ & =\frac {5}{p^{2}+p-2}-\frac {2p}{p^{2}+p-2}\nonumber \\ & =\frac {5}{\ \left (p-1\right ) \left (p+2\right ) }-\frac {2p}{\left ( p-1\right ) \left (p+2\right ) } \tag {1} \end {align}
From table, L7, we see that \(\mathcal {L}\left (\frac {e^{-at}-e^{-bt}}{b-a}\right ) =F\relax (p) =\frac {1}{\ \left (p+a\right ) \left (p+b\right ) }\). Setting \(a=-1,b=2\) we get the first expression in (1), that is
\begin {equation} \mathcal {L}\left (\frac {e^{t}-e^{-2t}}{3}\right ) =\frac {1}{\ \left (p-1\right ) \left (p+2\right ) } \tag {2} \end {equation}
also from table, we see that L8 is \(\mathcal {L}\left (\frac {ae^{-at}-be^{-bt}}{a-b}\right ) =F\relax (p) =\frac {p}{\ \left (p+a\right ) \left (p+b\right ) }\). Hence, letting \(a=-1,b=2\) we get the second expression in (1), that is
\begin {equation} \mathcal {L}\left (\frac {-e^{+t}-2e^{-2t}}{-3}\right ) =\frac {p}{\ \left (p-1\right ) \left (p+2\right ) } \tag {3} \end {equation}
So combine (2) and (3) we get
\[ 5\mathcal {L}\left (\frac {e^{+t}-e^{-2t}}{2+1}\right ) -2\mathcal {L}\left (\frac {-e^{+t}-2e^{-2t}}{-1-2}\right ) =\frac {5}{\ \left (p-1\right ) \left (p+2\right ) }-\frac {2p}{\left (p-1\right ) \left (p+2\right ) }\]
which is (1). Hence
\begin {align*} f\relax (t) & =5\frac {e^{+t}-e^{-2t}}{3}-2\frac {-e^{+t}-2e^{-2t}}{-3}\\ & =5\frac {e^{+t}-e^{-2t}}{3}+2\frac {-e^{+t}-2e^{-2t}}{3}\\ & =\frac {5e^{+t}-5e^{-2t}-2e^{+t}-4e^{-2t}}{3}\ \\ & =\frac {3e^{t}-9e^{-2t}}{3}\\ & =e^{t}-3e^{-2t} \end {align*}
Hence the inverse transform of \(F(p)\) is \(e^{t}-3e^{-2t}\)
Problem Use laplace transform to solve \(y^{\prime \prime }\ -4y=4e^{2t}\) ,\(y_{0}=0,y_{0}^{\prime }=1\)
Solution
Let \(\mathcal {L}\left (y\relax (t) \right ) =Y\relax (p) \), and taking the laplace transform of both sides, noting first that \(\mathcal {L}\left (y^{\prime \prime }\right ) =p^{2}Y-py_{0}-y_{0}^{\prime }\) then we get
\begin {align*} \mathcal {L}\left (y^{\prime \prime }\right ) \ -4\mathcal {L}\left (y\relax (t) \right ) & =\mathcal {L}\left (4e^{2t}\right ) \\ \left (p^{2}Y-py_{0}-y_{0}^{\prime }\right ) \ -4Y & =4\mathcal {L}\left (e^{2t}\right ) \end {align*}
L2 from table on page 636 : \(\mathcal {L}\left (e^{-at}\right ) =\frac {1}{p+a}\ \), hence \(\mathcal {L}\left (e^{2t}\right ) =\frac {1}{p-2}\), hence, after applying boundary conditions, we get
\begin {align*} \left (p^{2}Y\ -1\right ) \ -4Y & =4\frac {1}{p-2}\\ \ Y\left (p^{2}-4\right ) -1 & =4\frac {1}{p-2}\\ Y & =4\frac {1}{\left (p^{2}-4\right ) \left (p-2\right ) }+\frac {1}{\left (p^{2}-4\right ) }\\ Y & =\ 4\frac {1}{\left (p-2\right ) \left (p+2\right ) \left (p-2\right ) }+\frac {1}{\left (p^{2}-4\right ) }\\ Y & =\ 4\frac {1}{\ \left (p+2\right ) \left (p-2\right ) ^{2}}+\frac {1}{\left (p^{2}-4\right ) } \end {align*}
Doing partial fractions, repeated roots, gives
\begin {align*} \frac {1}{\ \left (p+2\right ) \left (p-2\right ) ^{2}} & =\frac {A}{\left ( p+2\right ) \ }+\frac {B}{\left (p-2\right ) \ }+\ \frac {C}{\left ( p-2\right ) ^{2}\ }\\ 1 & =A\left (p-2\right ) ^{2}+B\left (p+2\right ) \left (p-2\right ) +C\left (p+2\right ) \\ 1 & =A\left (p^{2}-4p+4\right ) +B\left (p^{2}-4\right ) +C\left ( p+2\right ) \\ 1 & =p^{2}\left (A+B\right ) +p\left (-4A+C\right ) +4A-4B+2C \end {align*}
Hence
\begin {align*} \ A+B & =0\\ -4A+C & =0\\ 4A-4B+2C & =1 \end {align*}
hence \(A=-B\), then \(-4B-4B+2C=1\) or \(-8B+2C=1\) or \(C=\frac {1+8B}{2}\), therefore \(4B+\frac {1+8B}{2}=0\) then \(B=-\frac {1}{16}\), then \(A=\frac {1}{16},C=\frac {1+8\left (-\frac {1}{16}\right ) }{2}=\frac {1-\ \left (\frac {1}{2}\right ) }{2}=\frac {1}{4}\). Hence
\begin {align*} Y & =4\frac {1}{\ \left (p+2\right ) \left (p-2\right ) ^{2}}+\frac {1}{\left (p^{2}-4\right ) }\\ & =4\left (\frac {A}{\left (p+2\right ) \ }+\frac {B}{\left (p-2\right ) \ }+\ \frac {C}{\left (p-2\right ) ^{2}\ }\right ) \ +\frac {1}{\left ( p^{2}-4\right ) }\\ & =4\left (\frac {\frac {1}{16}}{\left (p+2\right ) \ }+\frac {-\frac {1}{16}}{\left (p-2\right ) \ }+\ \frac {\frac {1}{4}}{\left (p-2\right ) ^{2}\ }\right ) \ +\frac {1}{\left (p^{2}-4\right ) }\\ & =\frac {1}{4}\frac {\ 1}{\left (p+2\right ) \ }-\frac {1}{4}\frac {1}{\left ( p-2\right ) \ }\ +\frac {1}{\left (p-2\right ) ^{2}}+\frac {1}{\left ( p^{2}-4\right ) }\\ & =\frac {1}{4}\frac {\ 1}{\left (p+2\right ) \ }-\frac {1}{4}\frac {1}{\left ( p-2\right ) \ }\ +\frac {1}{\left (p-2\right ) ^{2}}+\ \left (\frac {1}{4}\frac {\ 1}{\left (p-2\right ) \ }-\frac {1}{4}\frac {1}{\left (p+2\right ) \ }\right ) \\ & =\ \frac {1}{\left (p-2\right ) ^{2}} \end {align*}
\(\frac {1}{\left (p-2\right ) ^{2}}\rightarrow \) using L6, we have \(\mathcal {L}\left (\ t^{k}e^{-at}\right ) =\frac {k!}{\left (p+a\right ) ^{k+1}}\), here \(a=-2,k=1\) and \(\frac {1}{\left (p-2\right ) ^{2}}\rightarrow te^{2t}\). Hence ,
\[ f\relax (t) =te^{2t}\]
Problem Use laplace transform to solve \(y^{\prime \prime }\ -2y^{\prime }+y=2\cos t\) ,\(y_{0}=5,y_{0}^{\prime }=-2\)
Solution
Let \(\mathcal {L}\left (y\relax (t) \right ) =Y\relax (p) \), Take the laplace transform of both sides, noting first that \(\mathcal {L}\left (y^{\prime \prime }\right ) =p^{2}Y-py_{0}-y_{0}^{\prime }\) , \(\mathcal {L}\left (y^{\prime }\right ) =pY-y_{0}\)then we get
\begin {align*} \mathcal {L}\left (y^{\prime \prime }\right ) \ -2\mathcal {L}\left (y^{\prime }\right ) +\mathcal {L}\left (y\relax (t) \right ) & =\mathcal {L}\left (2\cos t\right ) \\ \left (p^{2}Y-py_{0}-y_{0}^{\prime }\right ) \ -2\left (pY-y_{0}\right ) +Y & =2\frac {p}{p^{2}+1} \end {align*}
\begin {align*} \left (p^{2}Y-5p+2\right ) \ -2\left (pY-5\right ) +Y & =2\frac {p}{p^{2}+1}\\ Y\left (p^{2}-2p+1\right ) +12-5p & =2\frac {p}{p^{2}+1}\\ Y\left (p^{2}-2p+1\right ) & =\frac {2p}{p^{2}+1}-12+5p\\ Y & =\frac {\frac {2p}{p^{2}+1}-12+5p}{\left (p^{2}-2p+1\right ) }\\ Y & =\ \ \frac {2p}{\left (p^{2}+1\right ) \left (p^{2}-2p+1\right ) }+\frac {5p-12}{\left (p^{2}-2p+1\right ) }\\ Y & =\ \frac {2p}{\left (p^{2}+1\right ) \left (p-1\right ) ^{2}}+\frac {5p-12}{\left (p-1\right ) ^{2}} \end {align*}
Doing partial fractions \(\frac {2p}{\left (p-1\right ) ^{2}\left ( p^{2}+1\right ) }=\frac {A}{\left (p-1\right ) }+\frac {B}{\left (p-1\right ) ^{2}\ }+\ \frac {Cp+D}{\left (p^{2}+1\right ) \ }\). Solving, we get \(A=0,B=1,C=0,D=-1\) Hence
\begin {align*} Y & =\ \frac {1}{\left (p-1\right ) ^{2}\ }-\ \frac {\ 1}{\left ( p^{2}+1\right ) \ }+\frac {5p-12}{\left (p-1\right ) ^{2}}\\ & =\ \frac {1}{\left (p-1\right ) ^{2}\ }-\ \frac {\ 1}{\left (p^{2}+1\right ) \ }+\frac {5p\ }{\left (p-1\right ) ^{2}}-12\frac {\ 1\ }{\left ( p-1\right ) ^{2}}\\ & =\frac {-11}{\left (p-1\right ) ^{2}\ }-\ \frac {\ 1}{\left (p^{2}+1\right ) \ }+\frac {5p\ }{\left (p-1\right ) ^{2}} \end {align*}
Hence , using table, we get inverse laplace transform \(\frac {1}{\left ( p-1\right ) ^{2}\ }\rightarrow te^{t}\) and \(\ \frac {\ 1}{\left ( p^{2}+1\right ) \ }\rightarrow \sin t\) and \(\frac {p\ }{\left (p-1\right ) ^{2}}\rightarrow e^{t}+te^{t}\), hence
\begin {align*} f\relax (t) & =-11te^{t}-\sin t+5\left (e^{t}+te^{t}\right ) \\ & =-11te^{t}-\sin t+5e^{t}+5te^{t}\\ & =5e^{t}-6te^{t}-\sin t \end {align*}
Problem Use laplace transform to solve \(y^{\prime \prime }\ +4y^{\prime }+5y=2e^{-2t}\cos t,\) ,\(y_{0}=0,y_{0}^{\prime }=3\)
Solution
Let \(\mathcal {L}\left (y\relax (t) \right ) =Y\relax (p) \), Taking the laplace transform of both sides, noting first that \(\mathcal {L}\left (y^{\prime \prime }\relax (t) \right ) =p^{2}Y-py_{0}-y_{0}^{\prime }\) , \(\mathcal {L}\left (y^{\prime }\relax (t) \right ) =pY-y_{0}\)then we get
\begin {align*} \mathcal {L}\left (y^{\prime \prime }\relax (t) \right ) +4\mathcal {L}\left (y^{\prime }\relax (t) \right ) +5\mathcal {L}\left (y\relax (t) \right ) & =\mathcal {L}\left (2e^{-2t}\cos t\right ) \\ \left (p^{2}Y-py_{0}-y_{0}^{\prime }\right ) \ +4\left (pY-y_{0}\right ) +5Y & =2\mathcal {L}\left (e^{-2t}\cos t\right ) \end {align*}
Applying initial conditions
\[ \left (p^{2}Y\ -3\right ) \ +4\left (pY\right ) +5Y=2\mathcal {L}\left (e^{-2t}\cos t\right ) \]
From table using L14 \(\mathcal {L}\left (e^{-2t}\cos t\right ) =\frac {p+2}{\left (p+2\right ) ^{2}+1}\). Hence
\begin {align*} \left (p^{2}Y\ -3\right ) \ +4\left (pY\right ) +5Y & =2\frac {p+2}{\left ( p+2\right ) ^{2}+1}\\ Y\left (p^{2}+4p+5\right ) -3 & =2\frac {p+2}{\left (p+2\right ) ^{2}+1}\\ Y & =\frac {2\frac {p+2}{\left (p+2\right ) ^{2}+1}+3}{\left (p^{2}+4p+5\right ) }\\ Y & =2\frac {p+2}{\left (\left (p+2\right ) ^{2}+1\right ) \left ( p^{2}+4p+5\right ) }+3\frac {1}{\left (p^{2}+4p+5\right ) } \end {align*}
but \(\left (p^{2}+4p+5\right ) =\left (p+2\right ) ^{2}+1\), therefore
\begin {align*} Y & =2\frac {p+2}{\left (\left (p+2\right ) ^{2}+1\right ) ^{2}\ }+\frac {3}{\left (p+2\right ) ^{2}+1}\\ & =2\frac {p+2}{\left (\left (p+2\right ) ^{2}+1\right ) \left (\left ( p+2\right ) ^{2}+1\right ) }+\frac {3}{\left (p+2\right ) ^{2}+1} \end {align*}
But inverse transform of \(\frac {p+2}{\left (\left (p+2\right ) ^{2}+1\right ) ^{2}\ }=\frac {1}{2}te^{-2t}\sin t\) and inverse transform of \(\frac {1}{\left (p+2\right ) ^{2}+1}=e^{-2t}\sin t\). Hence
\begin {align*} f\relax (t) & =2\left (\frac {1}{2}te^{-2t}\sin t\right ) +3\left ( e^{-2t}\sin t\right ) \\ & =te^{-2t}\sin t+3e^{-2t}\sin t\\ & =\left (t+3\right ) e^{-2t}\ \sin t \end {align*}
Problem Use laplace transform to solve \(y^{\prime }\ +z^{\prime }-2y=1,\) \(y_{0}=z_{0}=1,z-y^{\prime }=t\ \)
Solution
Taking laplace transform of both equations, then we get 2 equations in \(Y\) and \(Z\), then solve for them. Let \(\mathcal {L}\left (y\relax (t) \right ) =Y\relax (p) \), \(\mathcal {L}\left (z\relax (t) \right ) =Z\relax (p) \)
\begin {align*} \mathcal {L}\left (y^{\prime }\relax (t) \right ) \ +\mathcal {L}\left (z^{\prime }\relax (t) \right ) -2\mathcal {L}\left (y\relax (t) \right ) & =\mathcal {L}\relax (1) \\\mathcal {L}\left (z\relax (t) \right ) \ -\mathcal {L}\left (y^{\prime }\relax (t) \right ) & =\mathcal {L}\relax (t) \end {align*}
Then we get
\begin {align*} \left (pY-y_{0}\ \right ) +\left (pZ-z_{0}\ \right ) -2Y & =\frac {1}{p}\\ \ Z\ -\left (pY-y_{0}\ \right ) \ & =\ \frac {1}{p^{2}} \end {align*}
Putting initial conditions gives
\begin {align*} \left (pY-1\ \right ) +\left (pZ-1\ \right ) -2Y & =\frac {1}{p}\\ \ Z\ -\left (pY-1\ \right ) \ & =\ \frac {1}{p^{2}} \end {align*}
\begin {align} \ Y\left (p-2\right ) +\ pZ-2\ \ \ & =\frac {1}{p}\tag {1}\\ \ Z\ -pY+1\ \ & =\ \frac {1}{p^{2}} \tag {2} \end {align}
Solving for \(Y,\)From (1), \(Z=\frac {\frac {1}{p}+2-Y\left (p-2\right ) }{p}\) , and substituting into (2) gives
\begin {align*} \ \frac {\frac {1}{p}+2-Y\left (p-2\right ) }{p}-pY+1\ & =\ \frac {1}{p^{2}}\\ \frac {1}{p}+2-Y\left (p-2\right ) -p^{2}Y+p & =\frac {1}{p}\\ \ \ 2-Y\left (p-2\right ) -p^{2}Y\ & =-p\\ \ Y\left (-p+2-p^{2}\right ) \ & =\ -p-2\\ Y\ & =\ \ \frac {-p-2}{\left (-p+2-p^{2}\right ) }\\ Y\ & =\ \ -\ \frac {p+2}{\left (-p+1\right ) \left (p+2\right ) }\\ Y & =-\ \frac {1}{\left (-p+1\right ) } \end {align*}
Hence \(Y=\frac {1}{p-1}\) so from L2
\[ y\relax (t) =e^{t}\]
Now, that we have \(Y,\) we solve for \(Z.\) From (2)
\begin {align*} \ Z\ -pY+1\ \ & =\ \frac {1}{p^{2}}\\ Z-p\left (\frac {1}{p-1}\right ) +1 & =\frac {1}{p^{2}}\\ Z & =\frac {1}{p^{2}}-1+\frac {p}{p-1}\\ Z & =\ \frac {p-1-p^{2}\left (p-1\right ) +p^{3}}{p^{2}\left (p-1\right ) }\\ Z & =\frac {p-1\ +p^{2}\ }{p^{2}\left (p-1\right ) }\\ Z & =\frac {p-1\ +p^{2}\ }{p^{2}\left (p-1\right ) } \end {align*}
Doing partial fraction on the above, we get \(Z=\ \frac {1}{p^{2}}+\frac {1}{p-1}\), Hence
\[ z\relax (t) =t+e^{t}\]
Problem Use laplace transform to solve \(y^{\prime }\ +2z=1,\) \(y_{0}=0,\ 2y-z^{\prime }=2t,z_{0}=1\ \)
Solution
Take laplace transform of both equations, then we get 2 equations in \(Y\) and \(Z\), then solve for them.
Let \(\mathcal {L}\left (y\relax (t) \right ) =Y\relax (p) \), \(\mathcal {L}\left (z\relax (t) \right ) =Z\relax (p) \)
\begin {align*} \mathcal {L}\left (y^{\prime }\relax (t) \right ) +2Z & =\mathcal {L}\relax (1) \ \\ \ 2Y\ -\mathcal {L}\left (z^{\prime }\relax (t) \right ) & =\mathcal {L}\left (2t\right ) \end {align*}
\begin {align*} pY-y_{0}\ +2Z & =\mathcal {L}\relax (1) \ \\ \ 2Y\ -\left (pZ-z_{0}\right ) \ & =\mathcal {L}\left (2t\right ) \end {align*}
Then we get, by putting \(z_{0}=1,y_{0}=0\)
\begin {align} pY\ \ +2Z & =\frac {1}{p}\tag {1}\\ \ 2Y\ -\left (pZ-1\ \right ) \ & =\frac {2}{p^{2}} \tag {2} \end {align}
Obtain Z from first equation and sub into the second to solve for \(Y\), \(Z=\frac {\frac {1}{p}-pY}{2}\), Hence
\begin {align*} \ 2Y\ -\left (p\left (\frac {\frac {1}{p}-pY}{2}\right ) -1\ \right ) \ & =\frac {2}{p^{2}}\\ 2Y-\frac {p}{2}\left (\frac {1}{p}-pY\right ) +1 & =\ \frac {2}{p^{2}}\\ 2Y\ -\frac {p}{2}\left (\frac {1-p^{2}Y}{p}\right ) & =\ \ \frac {2}{p^{2}}-1\\ 2Y\ -\frac {1}{2}\left (1-p^{2}Y\right ) & =\ \frac {2-p^{2}}{p^{2}}\\ 2Y\ -\frac {1}{2}\ +\frac {1}{2}p^{2}Y & =\frac {2-p^{2}}{p^{2}}\\ Y\left (2+\frac {p^{2}}{2}\right ) & =\frac {2-p^{2}}{p^{2}}+\frac {1}{2}\\ Y & =\ \frac {\frac {4-p^{2}}{2p^{2}}}{\left (\ \frac {4+p^{2}}{2}\right ) }\\ Y & =\frac {\ 4-p^{2}}{\ p^{2}\left (\ 4+p^{2}\right ) } \end {align*}
Hence , using partial fraction gives
\begin {align*} Y & =\frac {\ 4+p^{2}}{\ p^{2}\left (\ 4+p^{2}\right ) }\\ & =\frac {Ap+B}{p^{2}}+\frac {Cp+D}{\left (\ 4+p^{2}\right ) } \end {align*}
Then
\(\ \)\begin {align*} \left (Ap+B\right ) \left (\ 4+p^{2}\right ) +\left (Cp+D\right ) p^{2} & =4-p^{2}\\ 4Ap+Ap^{3}+4B+Bp^{2}+Cp^{3}+Dp^{2} & =4-p^{2}\\ p^{3}\left (A+C\right ) +p^{2}\left (B+D\right ) +p\left (4A\right ) +4B & =4-p^{2} \end {align*}
Hence, \(4B=4\rightarrow B=1\ \) and \(4A=0\rightarrow A=0\ \) and \(B+D=-1\) and \(A+C=0\rightarrow C=0\ \), therefore \(D=-1-B\rightarrow D=-2\ \). Hence
\begin {align*} Y & =\frac {Ap+B}{p^{2}}+\frac {Cp+D}{\left (\ 4+p^{2}\right ) }\\ & =\frac {1}{p^{2}}-\frac {2}{\left (\ 4+p^{2}\right ) } \end {align*}
Using tables for inverse transform gives
\[ y\relax (t) =t-\sin 2t \]
Now, to find \(z\relax (t) \), subtituting value we found for Y into equation (1) above.
\begin {align*} pY\ \ +2Z & =\frac {1}{p}\\ p\left (\frac {1}{p^{2}}-\frac {2}{\left (\ 4+p^{2}\right ) }\right ) \ \ +2Z & =\frac {1}{p}\\ \frac {1}{p}-\frac {2p}{\left (\ 4+p^{2}\right ) }\ \ +2Z & =\frac {1}{p}\\ \ \ Z & =\frac {p}{\left (\ 4+p^{2}\right ) }\\ Z & =\frac {p}{\left (\ 4+p^{2}\right ) } \end {align*}
From tables, using L4
\[ z\relax (t) =\cos 2t \]
Problem Use laplace transform to solve \(y^{\prime \prime }+y=\sin t\) ,\(y_{0}=1,y_{0}^{\prime }=0\)
Solution
Let \(\mathcal {L}\left (y\relax (t) \right ) =Y\relax (p) \), Take the laplace transform of both sides, noting first that \(\mathcal {L}\left (y^{\prime \prime }\relax (t) \right ) =p^{2}Y-py_{0}-y_{0}^{\prime }\) then we get
\begin {align*} \mathcal {L}\left (y^{\prime \prime }\relax (t) \right ) +\mathcal {L}\left (y\relax (t) \right ) & =\mathcal {L}\left (\sin t\right ) \\ \left (p^{2}Y-py_{0}-y_{0}^{\prime }\right ) +Y\ & =\frac {1}{p^{2}+1} \end {align*}
Where I used L3 from table on page 636 which says that \(\mathcal {L}\left (\sin at\right ) =\frac {a}{p^{2}+a^{2}}\). Now solving for \(Y,\) noting that \(y_{0}=1\) and \(y_{0}^{\prime }=0\) gives
\begin {align} \left (p^{2}Y-p\right ) +Y\ & =\frac {1}{p^{2}+1}\nonumber \\ Y\left (p^{2}+1\right ) \ -p & =\frac {1}{p^{2}+1}\nonumber \\ Y\left (p^{2}+1\right ) \ \ & =\frac {1}{p^{2}+1}+p\ \nonumber \\ Y\ & =\ \frac {1}{\left (p^{2}+1\right ) ^{2}}+\ \frac {p}{\left ( p^{2}+1\right ) } \tag {1} \end {align}
From table using L12, \(\frac {p}{\left (p^{2}+1\right ) }\rightarrow \cos t\) and using L17, \(\frac {1}{\left (p^{2}+1\right ) ^{2}}\rightarrow \frac {1}{2}\left (\sin t-t\cos t\right ) \). Hence, putting these together into (1) gives
\[ f\relax (t) =\cos t+\frac {1}{2}\left (\sin t-t\cos t\right ) \]
This is the particular solution to the ODE.
Problem Use laplace transform to solve \(y^{\prime \prime }-6y^{\prime }+9y=te^{3t}\) ,\(y_{0}=0,y_{0}^{\prime }=5\)
Solution
Let \(\mathcal {L}\left (y\relax (t) \right ) =Y\relax (p) \), Take the laplace transform of both sides, noting first that \(\mathcal {L}\left (y^{\prime \prime }\relax (t) \right ) =p^{2}Y-py_{0}-y_{0}^{\prime }\) and \(\mathcal {L}\left (y^{\prime }\relax (t) \right ) =pY-y_{0}\) then we get
\begin {align*} \mathcal {L}\left (y^{\prime \prime }\relax (t) \right ) -6\mathcal {L}\left (y^{\prime }\relax (t) \right ) +9\mathcal {L}\left (y\relax (t) \right ) & =\mathcal {L}\left (t\ e^{3t}\right ) \\ \left (p^{2}Y-py_{0}-y_{0}^{\prime }\right ) -6\left (pY-y_{0}\right ) +9Y & =\mathcal {L}\left (t\ e^{3t}\right ) \end {align*}
I use L6 from table on page 636 which says that \(\mathcal {L}\left (t^{k}\ e^{-at}\right ) =\frac {k!}{\left (p+a\right ) ^{k+1}},\) hence for \(k=1,a=-3\), we get \(\mathcal {L}\left (t^{k}\ e^{-at}\right ) =\frac {1}{\left (p-3\right ) ^{2}}\)
\[ \left (p^{2}Y-py_{0}-y_{0}^{\prime }\right ) -6\left (pY-y_{0}\right ) +9Y=\frac {1}{\left (p-3\right ) ^{2}}\]
Applying boundary conditions gives
\begin {align} \left (p^{2}Y-5\right ) -6\left (pY\right ) +9Y & =\frac {1}{\left ( p-3\right ) ^{2}}\nonumber \\ Y\left (p^{2}-6p+9\right ) -5 & =\frac {1}{\left (p-3\right ) ^{2}}\nonumber \\ Y\left (p^{2}-6p+9\right ) \ & =\frac {1}{\left (p-3\right ) ^{2}}+5\nonumber \\ Y & =\frac {1}{\left (p^{2}-6p+9\right ) \left (p-3\right ) ^{2}}+\frac {5}{\left (p^{2}-6p+9\right ) }\nonumber \\ Y & =\frac {1}{\ \left (p-3\right ) ^{2}\left (p-3\right ) ^{2}}+\frac {5}{\left (p-3\right ) ^{2}}\nonumber \\ Y & =\frac {1}{\ \left (p-3\right ) ^{4}\ }+\frac {5}{\left (p-3\right ) ^{2}} \tag {1} \end {align}
Now using table, from L6, \(\frac {1}{\left (p-3\right ) ^{2}},\) let \(a=-3,k=1\) hence \(\frac {1}{\left (p-3\right ) ^{2}}\rightarrow te^{3t}\) And using L6 again, \(\frac {1}{\ \left (p-3\right ) ^{4}\ },\) let \(k=3,a=-3\) then \(\frac {6}{\ \left (p-3\right ) ^{4}\ }\rightarrow t^{3}e^{3t}\), therefore (1) becomes
\begin {align*} f\relax (t) & =\frac {1}{6}t^{3}e^{3t}+5\ te^{3t}\\ & =e^{3t}\left (\frac {1}{6}t^{3}+5t\right ) \end {align*}
Problem Use laplace transform to solve \(y^{\prime \prime }\ +16y=8\cos 4t\) ,\(y_{0}=0,y_{0}^{\prime }=0\)
Solution
Let \(\mathcal {L}\left (y\relax (t) \right ) =Y\relax (p) \), Taking the laplace transform of both sides, noting first that \(\mathcal {L}\left (y^{\prime \prime }\relax (t) \right ) =p^{2}Y-py_{0}-y_{0}^{\prime }\) results in
\begin {align*} \mathcal {L}\left (y^{\prime \prime }\relax (t) \right ) \ +16\mathcal {L}\left (y\relax (t) \right ) & =\mathcal {L}\left (8\cos 4t\right ) \\ \left (p^{2}Y-py_{0}-y_{0}^{\prime }\right ) \ +16Y & =8\mathcal {L}\left (\cos 4t\right ) \\ Y\left (p^{2}+16\right ) & =8\frac {p}{p^{2}+16} \end {align*}
I used L6 from table on page 636 : \(\mathcal {L}\left (\cos at\right ) =\frac {p}{p^{2}+a^{2}}\ \)
\[ Y=8\frac {p}{\left (p^{2}+16\right ) ^{2}}\]
Now looking at L11, which says \(\mathcal {L}\left (t\ \sin at\right ) =\frac {2ap}{\left (p^{2}+a^{2}\right ) ^{2}}\), hence letting \(a=4\,\ \) gives the solution
\[ \ f\relax (t) =t\ \sin 4t\ \]