Solving linear ODEs using the Laplace transform

Problem: Solve the ODE $\;\dfrac{df}{dt} + 2f = 0\;$ with the initial condition $\;f(0)=5$

Self-contained HTML rendered with MathJax (LaTeX)

Solution

Transform the ODE

Taking Laplace transforms of both sides and using linearity together with $\mathcal{L}\{f'(t)\}=sF(s)-f(0)$ gives \[ \mathcal{L}\{\tfrac{df}{dt}\}+2\,\mathcal{L}\{f\}=0 \;\Rightarrow\; sF(s)-f(0)+2F(s)=0. \]

Solve for $F(s)$

Rearranging, \[ (s+2)F(s)=f(0) \quad \Rightarrow \quad F(s)=\frac{f(0)}{s+2}. \]

Inverse transform

Apply $\mathcal{L}^{-1}$ and use linearity: \[ f(t)=\mathcal{L}^{-1}\{F(s)\} =\mathcal{L}^{-1}\!\left\{\frac{f(0)}{s+2}\right\} = f(0)\,\mathcal{L}^{-1}\!\left\{\frac{1}{s+2}\right\}. \]

Use a table of inverse transforms

Since $\mathcal{L}^{-1}\{\tfrac{1}{s-a}\}=e^{at}$, we have \[ \mathcal{L}^{-1}\!\left\{\frac{1}{s+2}\right\}=e^{-2t}. \] Hence \[ f(t)=f(0) e^{-2t}. \]

Apply the initial condition

With $f(0)=5$, \[ \boxed{\,f(t)=5e^{-2t}\,}. \]