The Laplace transform is useful in solving linear integro-differential equations.
Using the differentiation and integration properties of LT, each term in the integro-differential equation is transformed.
Initial conditions are automatically taken into account.
We solve the resulting algebraic equation in the s-domain.
We then convert the solution back to the time domain by using the inverse transform
Use the Laplace transform to solve the differential equation
\(\frac{d^{2} v(t)}{d t^{2}}+6 \frac{d
v(t)}{d t}+8 v(t)=2 u(t) \quad\) subject to \(v(0)=1, v^{\prime}(0)=-2\)
Solution:
\[\left[s^{2} V(s)-s v(0)-v^{\prime}(0)\right]+6[s
V(s)-v(0)]+8 V(s)=\frac{2}{s}\] Substituting \(v(0)=1, v^{\prime}(0)=-2\), \[s^{2} V(s)-s+2+6 s V(s)-6+8
V(s)=\frac{2}{s}\] or \[\left(s^{2}+6
s+8\right) V(s)=s+4+\frac{2}{s}=\frac{s^{2}+4 s+2}{s}\] Hence,
\[V(s)=\frac{s^{2}+4
s+2}{s(s+2)(s+4)}=\frac{A}{s}+\frac{B}{s+2}+\frac{C}{s+4}\]
\[\begin{gathered} A=\left.s V(s)\right|_{s=0}=\left.\frac{s^{2}+4 s+2}{(s+2)(s+4)}\right|_{s=0}=\frac{2}{(2)(4)}=\frac{1}{4} \\ B=\left.(s+2) V(s)\right|_{s=-2}=\left.\frac{s^{2}+4 s+2}{s(s+4)}\right|_{s=-2}=\frac{-2}{(-2)(2)}=\frac{1}{2} \\ C=\left.(s+4) V(s)\right|_{s=-4}=\left.\frac{s^{2}+4 s+2}{s(s+2)}\right|_{s=-4}=\frac{2}{(-4)(-2)}=\frac{1}{4} \end{gathered}\] Hence, \[V(s)=\frac{\frac{1}{4}}{s}+\frac{\frac{1}{2}}{s+2}+\frac{\frac{1}{4}}{s+4}\] By the inverse Laplace transform, \[v(t)=\frac{1}{4}\left(1+2 e^{-2 t}+e^{-4 t}\right) u(t)\]
Solve for the response \(y(t)\) in the following integrodifferential equation. \[\frac{d y}{d t}+5 y(t)+6 \int_{0}^{t} y(\tau) d \tau=u(t), \quad y(0)=2\] Solution:
Taking the Laplace transform of each term, we get \[[s Y(s)-y(0)]+5 Y(s)+\frac{6}{s} Y(s)=\frac{1}{s}\] Substituting \(y(0)=2\) and multiplying through by \(s\), \[Y(s)\left(s^{2}+5 s+6\right)=1+2 s\] or \[Y(s)=\frac{2 s+1}{(s+2)(s+3)}=\frac{A}{s+2}+\frac{B}{s+3}\]
where \[\begin{gathered} A=\left.(s+2) Y(s)\right|_{s=-2}=\left.\frac{2 s+1}{s+3}\right|_{s=-2}=\frac{-3}{1}=-3 \\ B=\left.(s+3) Y(s)\right|_{s=-3}=\left.\frac{2 s+1}{s+2}\right|_{s=-3}=\frac{-5}{-1}=5 \end{gathered}\] Thus, \[Y(s)=\frac{-3}{s+2}+\frac{5}{s+3}\] Its inverse transform is \[y(t)=\left(-3 e^{-2 t}+5 e^{-3 t}\right) u(t)\]