Laplace Transforms

Each entry is one integral. For $e^{at}$, $\int_0^\infty e^{-(s-a)t}\,dt = \tfrac{1}{s - a}$ when $s > a$ (proved below). The trigonometric entries follow from the same integral through Euler's formula: $\cos bt$ and $\sin bt$ are the real and imaginary parts of $e^{ibt}$, whose transform is $\tfrac{1}{s - ib} = \tfrac{s + ib}{s^2 + b^2}$. The transform weighs $f(t)$ by the decaying factor $e^{-st}$ and adds up the result; the figure shows this for $f(t) = t$ at $s = 1$, where the shaded area under $te^{-t}$ is $\mathcal L\{t\}(1) = \tfrac{1}{1^2} = 1$.

Why multiplication by $s$: integrating by parts moves the derivative off $y$ and onto $e^{-st}$, and differentiating $e^{-st}$ brings down the factor $-s$; the boundary term at $t = 0$ leaves $-y(0)$ behind (see the proof below). So every derivative in the equation becomes a power of $s$ times $Y$ plus known initial values.

Reading backward is legitimate because two continuous functions with the same transform are equal (Lerch's theorem), so each transform has only one continuous inverse. Constants must match exactly: $\tfrac{6}{s^4}$ is $\mathcal L\{t^3\}$ because $3! = 6$, while $\tfrac{1}{s^4}$ would be $\mathcal L\left\{\tfrac{t^3}{6}\right\}$.

The shift rule is one line: $\int_0^\infty e^{-st}e^{at}f(t)\,dt = \int_0^\infty e^{-(s - a)t}f(t)\,dt = F(s - a)$. Multiplying by an exponential in $t$ slides the transform in $s$. That is why completed squares appear: the transform of the damped oscillation $e^{-t}\cos 2t$ is the transform of $\cos 2t$ with $s$ replaced by $s + 1$. The repeated-factor entry is the shift rule applied to $\mathcal L\{t\} = \tfrac1{s^2}$.

Step 3 — $A = \tfrac43$ and $B = \tfrac53$.

Step 4 — invert with the $e^{at}$ entry: $y(t) = \tfrac43e^{t} + \tfrac53e^{-2t}$.

Step 5 — check: $y(0) = 3$, which matches $\lim_{s\to\infty} sY(s) = 3$ (the initial-value theorem).

Step 3 — split: $Y = \dfrac{s + 1}{(s + 1)^2 + 4} + \dfrac{2}{(s + 1)^2 + 4}$.

Step 4 — each piece is a table entry with $s$ replaced by $s + 1$, i.e. the shift $a = -1$: $y(t) = e^{-t}\cos 2t + e^{-t}\sin 2t = e^{-t}(\cos 2t + \sin 2t)$. Without the shift the inverse would be $\cos 2t + \sin 2t$; the shift multiplies it by $e^{-t}$.

Step 3 — compare the $s^2$ coefficients: $0 = A + C$, so $A = -1$.

Step 4 — invert: $y(t) = -1 + t + e^{-t}$.

Step 5 — check by transforming back: $-\tfrac1s + \tfrac{1}{s^2} + \tfrac{1}{s + 1} = \dfrac{-s(s + 1) + (s + 1) + s^2}{s^2(s + 1)} = \dfrac{1}{s^2(s + 1)}$ ✓.

Substitute $\mathcal L\{y''\} = s^2Y - sy_0 - y_1$ and $\mathcal L\{y'\} = sY - y_0$ and collect the $Y$ terms: $a(s^2Y - sy_0 - y_1) + b(sY - y_0) + cY = F$. The roots of $as^2 + bs + c$ are the characteristic roots, so the partial-fraction pieces of $Y$ over them reproduce the homogeneous solutions, while the poles of $F$ produce the particular part. A constant force $K$ switched on at $t = 0$ has $F = \tfrac Ks$, and its pole at $s = 0$ is the new equilibrium.

Step 3 — partial fractions: $s - 1 = A(s + 1) + B(s - 2)$; $s = 2$ gives $A = \tfrac13$ and $s = -1$ gives $B = \tfrac23$.

Step 4 — $y(t) = \tfrac13e^{2t} + \tfrac23e^{-t}$.

Step 5 — check: $y(0) = 1$ ✓ and $y'(0) = \tfrac23 - \tfrac23 = 0$ ✓, and both exponentials are homogeneous solutions, since $2$ and $-1$ are the characteristic roots.

Step 3 — $Y = \dfrac2s - \dfrac{3}{s + 1} + \dfrac{1}{s + 3}$, so $y(t) = 2 - 3e^{-t} + e^{-3t}$.

Step 4 — check: $y(0) = 2 - 3 + 1 = 0$ ✓, $y'(0) = 3 - 3 = 0$ ✓, and $y'' + 4y' + 3y = (-3e^{-t} + 9e^{-3t}) + 4(3e^{-t} - 3e^{-3t}) + 3(2 - 3e^{-t} + e^{-3t}) = 6$ ✓.

Step 5 — the system settles at the new equilibrium $y = 2$ (where $3y = 6$) without overshooting, because both characteristic roots are real.

Step 3 — complete the square: $\dfrac{s + 2}{(s + 1)^2 + 1} = \dfrac{s + 1}{(s + 1)^2 + 1} + \dfrac{1}{(s + 1)^2 + 1}$.

Step 4 — invert with the shift rule: $y(t) = 1 - e^{-t}(\cos t + \sin t)$.

Step 5 — the overshoot: $y'(t) = 2e^{-t}\sin t$ first vanishes again at $t = \pi$, where $y(\pi) = 1 + e^{-\pi} \approx 1.043$, about $4\%$ above the final value $1$. The complex roots $-1 \pm i$ are what make the response overshoot.

Iterating gives $\mathcal L\{y''\} = s^2\mathcal L\{y\} - sy(0) - y'(0)$ — initial conditions enter automatically, which is the method's superpower. $\blacksquare$

(2) Solve the resulting algebraic equation for $Y(s)$ — always possible, it is linear in $Y$. (3) Partial-fraction $Y$ into table-sized pieces: $\tfrac{1}{(s-a)(s-b)} = \tfrac{1}{a-b}\left(\tfrac{1}{s-a} - \tfrac{1}{s-b}\right)$, quadratic pieces aimed at the sine/cosine entries. (4) Invert with the table. The bookkeeping wins where classical methods strain: piecewise forcing (step functions), impulses, and long chains of initial conditions all ride through the algebra unchanged. The cost is fluency with a small table — and partial fractions, which do the heavy lifting in every problem.

So $Y = \tfrac1s - \tfrac{1}{s+1}$.

Step 4 — invert with the table: $y(t) = 1 - e^{-t}$.

Step 5 — verify: $y' = e^{-t}$ and $y' + y = e^{-t} + 1 - e^{-t} = 1$ ✓, $y(0) = 0$ ✓. Compare with the integrating-factor route: same answer, but here the initial condition entered at Step 1 and no constant $C$ ever appeared.

Step 1 — transform: $s^2Y - s\cdot0 - 3 + Y = 0$.

Step 2 — solve: $Y = \dfrac{3}{s^2 + 1}$.

Step 3 — no partial fractions needed: this is $3$ times the table entry $\mathcal L\{\sin t\} = \tfrac{1}{s^2+1}$.

Step 4 — invert: $y = 3\sin t$.

Step 5 — verify: $y'' = -3\sin t = -y$ ✓, $y(0) = 0$ ✓, $y'(0) = 3\cos 0 = 3$ ✓. Now the same skeleton with forcing, damping, or a step function only lengthens

Step 3 — the pipeline itself never changes shape, which is exactly its virtue.