Question 8.13: Solution of a differential equation with initial conditions ......
Solution of a differential equation with initial conditions using the unilateral Laplace transform
Solve the differential equation
\mathbf{x}^{\prime\prime}(t)+7\mathbf{x}^{\prime}(t)+12\mathbf{x}(t)=0
for times t > 0 subject to the initial conditions
{{\mathbf{x}}}(0^{-})=2\ and {\frac{d}{d t}}(\mathbf{x}(t))_{t=0^{-}}=-4.
First, Laplace transform both sides of the equation.
s^{2}\,X(s)-s\,X(0^{-})-{\frac{d}{d t}}({\bf x}(t))_{t=0^{-}}+7[s\,{\bf X}(s)-{\bf x}(0^{-})]+12\,{\bf X}(s)=0
Then solve for X(s).
\mathrm{X}(s)={\frac{s\mathrm{X}(0^{-})+7\mathrm{x}(0^{-})+{\frac{d}{d t}}(\mathrm{x}(t))_{t=0^-}}{s^{2}+7s+12}}
or
X(s)={\frac{2s+10}{s^{2}+7s+12}}.
Expanding X(s) in partial fractions,
X(s)={\frac{4}{s+3}}-{\frac{2}{s+4}}.
From the Laplace transform table,
e^{-\alpha t}\,\mathrm{u}(t)\mathrm{}^{}\overset{\mathcal L}{\longleftrightarrow } {\frac{1}{s+\alpha}}.
Inverse Laplace transforming ,x(t)=(4e^{-3t}-2e^{-4t})\mathrm{u}(t). Substituting this result into the original differential equation, for times t ≥ 0
\frac{d^{2}}{d t^{2}}[4e^{-3t}-2e^{-4t}]+7\frac{d}{d t}[4e^{-3t}-2e^{-4t}]+12[4e^{-3t}-2e^{-4t}]=0
36e^{-3t}-32e^{-4t}-84e^{-3t}+56e^{-4t}+48e^{-3t}-24e^{-4t}=0
0=0
proving that the x( t) found actually solves the differential equation. Also
x(0^{-})=4-2=2 and {\frac{d}{d t}}(x(t))_{t=0^{-}}=-12+8=-4
which verifies that the solution also satisfies the stated initial conditions.