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Dated: 30-10-2024

Theorems

The First Shift Theorem

If \(\mathscr L (F(t)) = f(s)\) then \(\mathscr L (e^{-at} F(t)) = f(s + a)\).

Example

\[\mathscr L (t^2) = \frac 2 {s^3}\]
\[\mathscr L (t^2 e^{4t}) = \frac 2 {(s - 4)^3}\]

Here \(a = -4\).

Multiplying by \(t\)

If \(\mathscr L (F(t)) = f(s)\) then

\[\mathscr L (F(t)) = - \frac d {ds} (f(s))\]

Example

\[\mathscr L (\sin 2t) = \frac 2 {s^2 + 4}\]
\[\mathscr L (t \sin 2t) = -\frac d {ds} \left(\frac 2 {s^2 + 4}\right) = \frac {4s} {(s^2 + 4)^2}\]

This theorem can be extended to following

\[\mathscr{L}(t^{n}F(t))=(-1)^{n}\frac{d^{n}}{ds^{n}}(f(s))\]

Dividing by \(t\)

If \(\mathscr L (F(t)) = f(s)\) then

\[\mathscr L \left(\frac {F(t)} t\right) = \int_s^\infty f(s) ds\]

Example

Determine

\[\mathscr L \left(\frac{\sin (at)}{t}\right) = \frac a {s^2 + a^2}\]
\[\mathscr{L}\left(\frac{\sin(at)}{t}\right)\]
\[=\int_{s}^{\infty}\frac{a}{s^{2}+a^{2}}ds\]
\[=\left[\tan^{-1}\left(\frac{s}{a}\right)\right]_{s}^{\infty}\]
\[=\frac{\pi}{2}-\tan^{-1}\left(\frac{s}{a}\right)\]
\[=\tan^{-1}\left(\frac{a}{s}\right)\]

Inverse Transforms

\[\mathscr L (4) = \frac 4 s\]
\[\implies \mathscr L ^{-1} \left(\frac 4 s\right) = 4\]

But what about

\[\mathscr L ^{-1}\left(\frac{3s}{s^2 - s - 6}\right) = \mathscr L ^ {-1}\left(\frac{1}{s+2} + \frac{1}{s - 3}\right)\]
\[= \mathscr L ^ {-1}\left(\frac{1}{s + 2}\right) + \mathscr L ^ {-1}\left(\frac{1}{s - 3}\right)\]
\[= e^{-2t} + 2 e^{3t}\]

Rules of Partial Fractions

  1. The degree of numerator should be lower than degree of denominator.
  2. Factorize the denominator into its prime factors. This determines the shapes of the partial fraction.
  3. A linear factor \((s + a)\) gives a partial fraction \(\frac A {s + a}\).
  4. A repeated factor \((s + a)^2\) gives a partial fraction \(\frac A {s + a} + \frac B {(s + a)^2}\).
  5. Similarly, \((s + a)^3\) gives a partial fraction \(\frac A {s + a} + \frac B {(s + a)^2} + \frac C {(s + a)^3}\).
  6. A quadratic factor \((s^2 + ps + q)\) gives \(\frac{Ps + Q}{s^2 + ps + q}\).
  7. Similarly, \((s^2 + ps + q)^2\) gives \(\frac{Ps + Q}{s^2 + ps + q} + \frac{Rs + T}{(s^2 + ps + q)^2}\).
  8. \(\frac{s + a}{(s + b) (s + c)}\) has gives \(\frac{A}{s + b} + \frac{B}{s + c}\).
  9. \(\frac{s + a}{(s + b) (s + c)^2}\) has gives \(\frac{A}{s + b} + \frac{B}{s + c} + \frac{C}{(s + c)^2}\).

Example

Solve the following equation given that \(x(0) = 5\) and \(\frac{dx}{dt}\bigg|_{t = 0} = 7\)

\[\frac{d^2x}{dt^2} - 3 \frac{dx}{dt} + 2x = 2e^{3t}\]
\[x^{\prime \prime}(t) - 3x^\prime(t) + 2x(t) = 2e^{3t}\]

Taking the laplace transform1 on both sides

\[\mathscr L (x^{\prime \prime}(t) - 3x^\prime(t) + 2x(t)) = \mathscr L (2e^{3t})\]
\[\mathscr {L}(x^{\prime\prime}(t))-3\mathscr{L}(x^{\prime}(t))+2\mathscr{L}(x(t))=2\mathscr{L}(e^{3t})\]
\[s^2\mathscr L (x(t)) - sx(0) - x^\prime(0) - 3\left(s \mathscr L (x(t)) - x(0)\right) + 2 \mathscr L (x(t)) = \frac 2 {(s - 3)}\]

Plugging the values in, we get

\[s^2 \mathscr L (x(t)) - s(7) - 7 - 3 \left(s \mathscr L (x(t)) + 3 (5)\right) + 2 \mathscr L (x(t)) = \frac 2 {s - 3}\]
\[s^{2}\mathscr{L}(x(t))-3s\mathscr{L}(x(t))+2\mathscr{L}(x(t))=\frac{2}{s-3}-8+5s\]
\[(s^{2}-3s+2)\mathscr{L}(x(t))=\frac{2-8s+24+5s^{2}-15s}{s-3}\]
\[\mathscr{L}(x(t))=\frac{5s^{2}-23s+24}{(s-1)(s-2)(s-3)}\]

After solving the partial fraction, the coefficients we get are \(A = 3\), \(B = 2\) and \(C = 0\)

\[\mathscr{L}(x(t))=\frac{3}{s-1}+\frac{2}{s-2}+\frac{0}{s-3}\]
\[\mathscr{L}(x(t))=\frac{3}{s-1}+\frac{2}{s-2}\]
\[x(t)=\mathscr{L}^{-1}\left(\frac{3}{s-1}\right)+\mathscr{L}^{-1}\left(\frac{2}{s-2}\right)=3e^{t}+2e^{2t}\]

References

Read more about notations and symbols.

  1. Read more about laplace transform. ↩