Exercises — Laplace of derivatives — key property for solving ODEs

$\mathcal{L}\{f^{\prime }(t)\}=sF(s)-f(0)$ $\mathcal{L}\{f^{\prime \prime }(t)\}=s^{2}F(s)-sf(0)-f^{\prime }(0)$ Read the shape: one factor of $s$ per prime removed, and each derivative you remove "pays a fee" of an initial value. For $f^{\prime \prime }$ the two fees are $sf(0)$ (higher $s$-power, lower derivative) and $f^{\prime }(0)$ (lower $s$-power, higher derivative).

The general rule is $\mathcal{L}\{f^{(n)}\}=s^{n}F(s)-s^{n-1}f(0)-s^{n-2}{f}^{\prime }(0)-\cdots -f^{(n-1)}(0).$ The term carrying $f^{(k)}(0)$ is $-s^{n-1-k}{f}^{(k)}(0)$, so the coefficient is $\boxed{-s^{n-1-k}}$. Check with $n=2$: $k=0$ gives $-s^1$ (the $-sf(0)$ term); $k=1$ gives $-s^0=-1$ (the $-f^{\prime }(0)$ term). ✓

Way A — differentiate first. $f^{\prime }(t)=2e^{2t}$, so $\mathcal{L}\{f^{\prime }\}=\frac{2}{s-2}$. Way B — use the rule. $f(0)=e^0=1$, so $\mathcal{L}\{f^{\prime }\}=sF(s)-f(0)=\frac{s}{s-2}-1=\frac{s-(s-2)}{s-2}=\frac{2}{s-2}.$ Both give $\frac{2}{s-2}$. ✓ The $-f(0)=-1$ was essential — without it we'd get $\frac{s}{s-2}$, which is wrong.

$f(0)=\cos 0=1$. Rule: $\mathcal{L}\{f^{\prime }\}=sF(s)-f(0)=\frac{s^2}{s^2+9}-1=\frac{s^2-(s^2+9)}{s^2+9}=\frac{-9}{s^2+9}.$ Direct check: $f^{\prime }(t)=-3\sin 3t$, and $\mathcal{L}\{-3\sin 3t\}=-3\cdot \frac{3}{s^2+9}=\frac{-9}{s^2+9}$. ✓

Here $F(s)=\frac{2}{s^2+4}$, $f(0)=\sin 0=0$, $f^{\prime }(t)=2\cos 2t$ so $f^{\prime }(0)=2$. $\mathcal{L}\{f^{\prime \prime }\}=s^{2}F(s)-sf(0)-f^{\prime }(0)=\frac{2s^2}{s^2+4}-0-2=\frac{2s^2-2(s^2+4)}{s^2+4}=\frac{-8}{s^2+4}.$ Direct check: $f^{\prime \prime }(t)=-4\sin 2t$, $\mathcal{L}\{-4\sin 2t\}=-4\cdot \frac{2}{s^2+4}=\frac{-8}{s^2+4}$. ✓

The boundary term is $-f(0)$, so it vanishes precisely when $\boxed{f(0)=0}$. Example: $f(t)=\sin t$ has $f(0)=0$, so $\mathcal{L}\{f^{\prime }\}=sF(s)=\frac{s}{s^2+1}$. Check directly: $f^{\prime }=\cos t$, $\mathcal{L}\{\cos t\}=\frac{s}{s^2+1}$. ✓ Insight: the "clean" formula is not universally wrong — it is a special case that happens exactly at the origin-crossing. It fails the instant your function starts away from zero.

The boundary term in the integration by parts is $[e^{-st}f(t){]}_0^{\infty }$. Look at the figure: the orange curve $e^{-st}$ crushes downward exponentially, while the blue $f(t)$ grows at most like $e^{ct}$. Their product $e^{-st}f(t)$ is bounded by $M{e}^{-(s-c)t}$ (the green envelope), which $\to 0$ as $t\to \infty$ whenever $s>c$. So the top limit contributes $0$. Only the bottom limit $t=0$ remains, giving $-e^{0}f(0)=-f(0)$. This is the entire reason the formula reads $sF(s)-f(0)$ and not something with a surviving tail. If $s\le c$ the integral would diverge and the transform wouldn't exist — see Exponential Order and Convergence of Integrals.

Let $g(t)=f^{\prime }(t)$. The first rule applies to any nice function, so $\mathcal{L}\{g^{\prime }\}=s\mathcal{L}\{g\}-g(0).$ Now $g^{\prime }=f^{\prime \prime }$, $\mathcal{L}\{g\}=\mathcal{L}\{f^{\prime }\}=sF(s)-f(0)$, and $g(0)=f^{\prime }(0)$. Substitute: $\mathcal{L}\{f^{\prime \prime }\}=s(sF(s)-f(0))-f^{\prime }(0)=s^{2}F(s)-sf(0)-f^{\prime }(0).$ ✓ Why it works: the rule "removes one prime for one $s$ and one fee." Removing two primes just runs the same move twice — no new integration needed. That is the inductive engine behind the general $s^{n}F(s)-\cdots$ pattern.

Transform: $(sY-y(0))-2Y=0\Rightarrow (s-2)Y=y(0)=5$. Solve: $Y=\frac{5}{s-2}$. Invert: ${\mathcal{L}}^{-1}\{\frac{1}{s-2}\}=e^{2t}$, so $\boxed{y(t)=5e^{2t}}$. Check: $y^{\prime }=10e^{2t}$, $y^{\prime }-2y=10e^{2t}-10e^{2t}=0$ ✓; $y(0)=5$ ✓.

Transform: $(s^{2}Y-sy(0)-y^{\prime }(0))+4Y=0$. Insert ICs: $s^{2}Y-s-0+4Y=0\Rightarrow (s^2+4)Y=s$. Solve: $Y=\frac{s}{s^2+4}$. Invert: this is exactly $\mathcal{L}\{\cos 2t\}$, so $\boxed{y(t)=\cos 2t}$. Check: $y^{\prime \prime }=-4\cos 2t=-4y$, so $y^{\prime \prime }+4y=0$ ✓; $y(0)=1$ ✓, $y^{\prime }(0)=-2\sin 0=0$ ✓.

Transform: $\mathcal{L}\{6\}=\frac{6}{s}$, so $(sY-0)+2Y=\frac{6}{s}\Rightarrow (s+2)Y=\frac{6}{s}$. $Y=\frac{6}{s(s+2)}.$ Partial fractions: $\frac{6}{s(s+2)}=\frac{A}{s}+\frac{B}{s+2}$. Multiply out: $6=A(s+2)+Bs$. Set $s=0$: $6=2A\Rightarrow A=3$. Set $s=-2$: $6=-2B\Rightarrow B=-3$. $Y=\frac{3}{s}-\frac{3}{s+2}.$ Invert: $\boxed{y(t)=3-3e^{-2t}}$. Check: $y^{\prime }=6e^{-2t}$, $y^{\prime }+2y=6e^{-2t}+6-6e^{-2t}=6$ ✓; $y(0)=3-3=0$ ✓; equilibrium $y\to 3$ as $t\to \infty$ (matches RHS/coeff $=6/2=3$) ✓.

Transform each term using the rules:

• $\mathcal{L}\{y^{\prime \prime }\}=s^{2}Y-sy(0)-y^{\prime }(0)=s^{2}Y-2s-3$
• $\mathcal{L}\{-3y^{\prime }\}=-3(sY-y(0))=-3sY+6$
• $\mathcal{L}\{2y\}=2Y$

Sum $=0$: $(s^{2}Y-2s-3)+(-3sY+6)+2Y=0\Rightarrow (s^2-3s+2)Y=2s-3.$ Factor: $s^2-3s+2=(s-1)(s-2)$, so $Y=\frac{2s-3}{(s-1)(s-2)}$. Partial fractions: $\frac{2s-3}{(s-1)(s-2)}=\frac{A}{s-1}+\frac{B}{s-2}$. Then $2s-3=A(s-2)+B(s-1)$. Set $s=1$: $-1=A(-1)\Rightarrow A=1$. Set $s=2$: $1=B(1)\Rightarrow B=1$. $Y=\frac{1}{s-1}+\frac{1}{s-2}\Rightarrow \boxed{y(t)=e^t+e^{2t}}.$ Check: $y^{\prime }=e^t+2e^{2t}$, $y^{\prime \prime }=e^t+4e^{2t}$; $y^{\prime \prime }-3y^{\prime }+2y=(1-3+2)e^t+(4-6+2)e^{2t}=0$ ✓; $y(0)=1+1=2$ ✓; $y^{\prime }(0)=1+2=3$ ✓.

Transform: $s^{2}Y-sy(0)-y^{\prime }(0)+Y=0\Rightarrow s^{2}Y-0-2+Y=0\Rightarrow (s^2+1)Y=2$. $Y=\frac{2}{s^2+1}=2\cdot \frac{1}{s^2+1}\Rightarrow \boxed{y(t)=2\sin t}.$ Tracing the ICs: the $-sy(0)$ term vanished because $y(0)=0$ — that removed any $\cos t$ piece. The $-y^{\prime }(0)=-2$ term is the sole survivor, and it produced the $\sin t$ amplitude $2$. So the slope IC alone shaped this answer. Check: $y^{\prime \prime }=-2\sin t=-y$ ✓; $y(0)=0$ ✓; $y^{\prime }(0)=2\cos 0=2$ ✓.

Transform: $s^{2}Y-sy(0)-y^{\prime }(0)=0\Rightarrow s^{2}Y-s-4=0\Rightarrow Y=\frac{s+4}{s^2}=\frac{1}{s}+\frac{4}{s^2}.$ Invert: ${\mathcal{L}}^{-1}\{\frac{1}{s}\}=1$, ${\mathcal{L}}^{-1}\{\frac{1}{s^2}\}=t$, so $\boxed{y(t)=1+4t}.$ Check: $y^{\prime }=4$, $y^{\prime \prime }=0$ ✓; $y(0)=1$ ✓; $y^{\prime }(0)=4$ ✓. Limiting insight: for $y^{\prime \prime }+ky=0$ the frequency is $\sqrt{k}$. As $k\to 0$, $\sin (\sqrt{k}t)/\sqrt{k}\to t$ and $\cos (\sqrt{k}t)\to 1$, so the oscillation "unrolls" into the straight line $y(0)+y^{\prime }(0)t$. The $k=0$ case is the degenerate limit where the two initial conditions map straight to the intercept and slope of a line — no oscillation at all.

Every solution above passed three checks: the ODE itself, the value IC, and (for 2nd order) the slope IC. If any check fails, you almost certainly dropped a fee term $-s^{n-1-k}{f}^{(k)}(0)$.