Proposition: Suppose \(y(t)\) is continuous on \([0,\infty )\) and of exponential order \(s_{0}\), and \(y'\) is continuous on \([0,\infty )\). Then \(\mathcal {L}(y)\) and \(\mathcal {L}(y')\) are defined for \(s>s_{0}\) and \(\mathcal {L}(y')=s\mathcal {L}(y)-y(0)\).

Proof: By integration by parts,

\(\displaystyle \mathcal {L}(y')=\lim _{T\rightarrow \infty }\int _{0}^{T}e^{-st}y'(t)\,dt\) \(\displaystyle =\lim _{T\rightarrow \infty } \left .e^{-st}y(t)\right |_{0}^{T}+s\int _{0}^{T}e^{-st}y(t)\,dt\).

Since \(y\) is of exponential order \(s_{0}\), for \(s>s_{0}\) we have \(\displaystyle \lim _{T\rightarrow \infty }e^{-sT}y(T)=0\) and \(\displaystyle \mathcal {L}(y)=\int _{0}^{\infty }e^{-st}y(t)\,dt\) converges. Hence \(\mathcal {L}(y')\) converges for \(s>s_{0}\) and \( \mathcal {L}(y')=-y(0)+s\mathcal {L}(y)\mbox { }_{\square }\)

Let’s use this to solve an IVP using Laplace transforms! Consider the first order IVP \(y'-2y=te^t\), \(y(0)=0\).

We take the Laplace transform of both sides of the equation.

By clearing fractions we get that \(1=A(s-2)(s-1)-(s-2)+(s-1)^2\). Evaluated at \(s=0\), \(1=2A+3\), implying \(A=-1\).

So \(Y=\mathcal {L}(y)=-\frac {1}{s-1}-\frac {1}{(s-1)^2}+\frac {1}{s-2}\) is the Laplace transform of the solution to this IVP. Hence \(y=\mathcal {L}^{-1} (Y)=-e^t-te^t+e^{2t}=-(t+1)e^t+e^{2t}\) is the solution to this IVP.

Well, if \(\mathcal {L}(y')=s\mathcal {L}(y)-y(0)\) it follows that \(\mathcal {L}(y'')=s\mathcal {L}(y')-y'(0)\) and hence
\(\mathcal {L}(y'')=s(s\mathcal {L}(y)-y(0))-y'(0)\).

So \(\mathcal {L}(y'')=s^2 \mathcal {L}(y)-y(0) \,s -y'(0)\).

Similarly \(\mathcal {L}(y''')=s^3 \mathcal {L}(y)-y(0) \,s^2 -y'(0)\,s-y''(0)\).

Let’s solve the IVP \(y''+4y'-5y=6e^{t}\mbox {, }y(0)=1\mbox {, }y'(0)=0\) using Laplace transforms. Let \(Y=\mathcal {L}(y)\).

Then \(s^2 Y-s-0+4(sY-1)-5Y=\frac {6}{s-1}\), equivalently, \((s^2+4s-5)Y=\frac {6}{s-1}+s+4\).

Therefore \(\displaystyle Y(s)=\frac {\frac {6}{s-1}+s+4}{s^{2}+4s-5}\) is the Laplace transform of the solution.

Simplifying this yields \(Y(s)=\dfrac {6+(s+4)(s-1)}{(s-1)^{2}(s+5)}=\dfrac {s^{2}+3s+2}{(s-1)^{2}(s+5)}\).

We can begin our PFD by having \(Y(s)=\dfrac {s^{2}+3s+2}{(s-1)^{2}(s+5)}=\dfrac {A}{s-1}+\dfrac {1}{(s-1)^{2}}+\dfrac {1/3}{s+5}\).

By clearing fractions,

So \(A=2/3\) and \(Y(s)=\dfrac {2/3}{s-1}+\dfrac {1}{(s-1)^{2}}+\dfrac {1/3}{s+5}\).

Then \(\displaystyle y(t)=(2/3)\mathcal {L}^{-1}\left (\frac {1}{s-1}\right )+\mathcal {L}^{-1}\left (\frac {1}{(s-1)^{2}}\right )+(1/3)\mathcal {L}^{-1}\left (\frac {1}{s+5}\right )\) \(\displaystyle =\frac {2}{3}e^{t}+te^{t}+\frac {1}{3}e^{-5t}\).

Let’s solve \(y''+y=\sin {(t)}\), \(y(0)=1\), \(y'(0)=2\) using Laplace transforms.

Let \(Y=\mathcal {L}(y)\). Then \(s^2Y-s-2+Y=\frac {1}{s^2+1}\) or equivalently,
\(\displaystyle (s^2+1)Y=s+2+\frac {1}{s^2+1}\).

Hence \(\displaystyle Y=\frac {s+2}{s^2+1}+\frac {1}{(s^2+1)^2}=\frac {s}{s^2+1}+2\frac {1}{s^2+1}+\frac {1}{(s^2+1)^2}\).

Dealing with the term \(\frac {1}{(s^2+1)^2}\) requires some finesse. Based on the Laplace table (see appendix) we re-express it as

\(\displaystyle \frac {1}{(s^2+1)^2}=\frac {0.5(s^2+1)-0.5(s^2-1)}{(s^2+1)^2}=\frac {1}{2}\,\frac {1}{s^2+1}-\frac {1}{2}\,\frac {s^2-1}{(s^2+1)^2}\).

Thus \(\displaystyle Y=\frac {s}{s^2+1}+\frac {5}{2}\,\frac {1}{s^2+1}-\frac {1}{2}\,\frac {s^2-1}{(s^2+1)^2}\).

Therefore the inverse Laplace transform,
\(y(t)=\mathcal {L}^{-1}(Y)=\cos {(t)}+\frac {5}{2}\sin {(t)}-\frac {1}{2}t\cos {(t)}\) is the IVP solution.