MATH 2410 — Differential Equations
Lecture Notes. This is NOT meant to be a substitute for coming to lecture or consulting the textbook directly, but rather a helpful resource to review for exams/ review what was discussed in lecture.
Lecture 16: Section 7.1 - Introduction to Laplace Transform ↑ Back to top
In Unit 3, we introduce the Laplace Transform and develop methods to solve initial value problems. We begin by defining what a transform is and then build toward the Laplace Transform.
Definition.
A transform \(T\) is an operation that converts a function into another function.
A transform \(T\) is an operation that converts a function into another function.
Common examples include differentiation and integration:
\[ x^n \mapsto nx^{n-1}, \quad e^x \mapsto e^x. \]
Remark.
Transforms are linear. For constants \(c_1, c_2\) and functions \(f, g\): \[ T(c_1f \pm c_2g) = c_1T(f) \pm c_2T(g). \]
Transforms are linear. For constants \(c_1, c_2\) and functions \(f, g\): \[ T(c_1f \pm c_2g) = c_1T(f) \pm c_2T(g). \]
Generalized Integral Transform
Consider a function of two variables \(f(x,y)\):
\[ \int_a^b f(x,y)\,dx = F(b,y) - F(a,y), \]This transforms a function of two variables into a function of one variable.
Example 1
Let \(f(x,y) = 2xy^2.\) \[ \int_1^3 2xy^2\,dx = y^2 \int_1^3 2x\,dx \] \[ = y^2[x^2]_1^3 = y^2(9 - 1) = 8y^2. \]
Let \(f(x,y) = 2xy^2.\) \[ \int_1^3 2xy^2\,dx = y^2 \int_1^3 2x\,dx \] \[ = y^2[x^2]_1^3 = y^2(9 - 1) = 8y^2. \]
To build the Laplace Transform:
- Start with \[ \int_a^b K(s,t)f(t)\,dt \]
- Extend to infinity: \[ F(s) = \int_0^\infty K(s,t)f(t)\,dt \]
We choose the function \(K(s,t) = e^{-st}\) since it decays to 0 faster than any polynomial decay.
Definition.
The Laplace Transform of \(f(t)\) is \[ \mathcal{L}\{f(t)\} = \int_0^\infty e^{-st}f(t)\,dt, \] defined for values of \(s\) where the integral converges.
The Laplace Transform of \(f(t)\) is \[ \mathcal{L}\{f(t)\} = \int_0^\infty e^{-st}f(t)\,dt, \] defined for values of \(s\) where the integral converges.
Example 2
Compute the Laplace Transform of a constant \(k\). \[ \mathcal{L}\{k\} = \int_0^\infty e^{-st}k\,dt = k\int_0^\infty e^{-st}\,dt \] \[ = k \left[ \frac{-1}{s}e^{-st} \right]_0^\infty \] \[ = k\lim_{b \to \infty}\left[ \frac{-1}{s}e^{-sb} + \frac{1}{s}e^{-s(0)} \right] \] As \(b \to \infty\), \(e^{-sb} \to 0\) (for \(s>0\)): \[ \mathcal{L}\{k\} = \frac{k}{s}, \quad s>0. \]
Compute the Laplace Transform of a constant \(k\). \[ \mathcal{L}\{k\} = \int_0^\infty e^{-st}k\,dt = k\int_0^\infty e^{-st}\,dt \] \[ = k \left[ \frac{-1}{s}e^{-st} \right]_0^\infty \] \[ = k\lim_{b \to \infty}\left[ \frac{-1}{s}e^{-sb} + \frac{1}{s}e^{-s(0)} \right] \] As \(b \to \infty\), \(e^{-sb} \to 0\) (for \(s>0\)): \[ \mathcal{L}\{k\} = \frac{k}{s}, \quad s>0. \]
Example 3
Compute the Laplace Transform of \(f(t)=mt+b\). Using linearity: \[ \mathcal{L}\{mt+b\} = m\mathcal{L}\{t\} + \mathcal{L}\{b\} \] Compute \(\mathcal{L}\{t\}\): \[ \mathcal{L}\{t\} = \int_0^\infty te^{-st}\,dt \] Integration by parts: Let \(u=t\), \(dv=e^{-st}dt\) \[ du=dt, \quad v=\frac{-1}{s}e^{-st} \] \[ \mathcal{L}\{t\} = \left[ \frac{-t}{s}e^{-st} \right]_0^\infty + \frac{1}{s}\int_0^\infty e^{-st}\,dt \] \[ = 0 + \frac{1}{s}\cdot \frac{1}{s} \] \[ = \frac{1}{s^2} \] Also: \[ \mathcal{L}\{b\} = \frac{b}{s} \] Thus: \[ \mathcal{L}\{mt+b\} = m\left(\frac{1}{s^2}\right) + \frac{b}{s} = \frac{m}{s^2} + \frac{b}{s}, \quad s>0. \]
Compute the Laplace Transform of \(f(t)=mt+b\). Using linearity: \[ \mathcal{L}\{mt+b\} = m\mathcal{L}\{t\} + \mathcal{L}\{b\} \] Compute \(\mathcal{L}\{t\}\): \[ \mathcal{L}\{t\} = \int_0^\infty te^{-st}\,dt \] Integration by parts: Let \(u=t\), \(dv=e^{-st}dt\) \[ du=dt, \quad v=\frac{-1}{s}e^{-st} \] \[ \mathcal{L}\{t\} = \left[ \frac{-t}{s}e^{-st} \right]_0^\infty + \frac{1}{s}\int_0^\infty e^{-st}\,dt \] \[ = 0 + \frac{1}{s}\cdot \frac{1}{s} \] \[ = \frac{1}{s^2} \] Also: \[ \mathcal{L}\{b\} = \frac{b}{s} \] Thus: \[ \mathcal{L}\{mt+b\} = m\left(\frac{1}{s^2}\right) + \frac{b}{s} = \frac{m}{s^2} + \frac{b}{s}, \quad s>0. \]
Example 4
Compute the Laplace Transform of \[ f(t)= \begin{cases} t, & 0 \le t \le 1\\ 1, & t \ge 1 \end{cases} \] Solution: \[ \mathcal{L}\{f(t)\} = \int_0^1 te^{-st}\,dt + \int_1^\infty e^{-st}\,dt \] First integral: Let \(u=t\), \(dv=e^{-st}dt\) \[ v=\frac{-1}{s}e^{-st} \] \[ \int_0^1 te^{-st}\,dt = \left[ \frac{-t}{s}e^{-st} \right]_0^1 + \frac{1}{s}\int_0^1 e^{-st}\,dt \] \[ = \left(\frac{-1}{s}e^{-s}\right) + \frac{1}{s}\left[\frac{-1}{s}e^{-st}\right]_0^1 \] \[ = \frac{-1}{s}e^{-s} + \frac{1}{s}\left(\frac{-1}{s}e^{-s} + \frac{1}{s}\right) \] \[ = \frac{-1}{s}e^{-s} - \frac{1}{s^2}e^{-s} + \frac{1}{s^2} \] Second integral: \[ \int_1^\infty e^{-st}\,dt = \left[\frac{-1}{s}e^{-st}\right]_1^\infty \] \[ = 0 + \frac{1}{s}e^{-s} \] Add: \[ \mathcal{L}\{f(t)\} = \left(\frac{-1}{s}e^{-s} - \frac{1}{s^2}e^{-s} + \frac{1}{s^2}\right) + \frac{1}{s}e^{-s} \] \[ = \frac{1}{s^2} - \frac{1}{s^2}e^{-s} \] Factor: \[ \mathcal{L}\{f(t)\} = \frac{1 - e^{-s}}{s^2}, \quad s>0. \]
Compute the Laplace Transform of \[ f(t)= \begin{cases} t, & 0 \le t \le 1\\ 1, & t \ge 1 \end{cases} \] Solution: \[ \mathcal{L}\{f(t)\} = \int_0^1 te^{-st}\,dt + \int_1^\infty e^{-st}\,dt \] First integral: Let \(u=t\), \(dv=e^{-st}dt\) \[ v=\frac{-1}{s}e^{-st} \] \[ \int_0^1 te^{-st}\,dt = \left[ \frac{-t}{s}e^{-st} \right]_0^1 + \frac{1}{s}\int_0^1 e^{-st}\,dt \] \[ = \left(\frac{-1}{s}e^{-s}\right) + \frac{1}{s}\left[\frac{-1}{s}e^{-st}\right]_0^1 \] \[ = \frac{-1}{s}e^{-s} + \frac{1}{s}\left(\frac{-1}{s}e^{-s} + \frac{1}{s}\right) \] \[ = \frac{-1}{s}e^{-s} - \frac{1}{s^2}e^{-s} + \frac{1}{s^2} \] Second integral: \[ \int_1^\infty e^{-st}\,dt = \left[\frac{-1}{s}e^{-st}\right]_1^\infty \] \[ = 0 + \frac{1}{s}e^{-s} \] Add: \[ \mathcal{L}\{f(t)\} = \left(\frac{-1}{s}e^{-s} - \frac{1}{s^2}e^{-s} + \frac{1}{s^2}\right) + \frac{1}{s}e^{-s} \] \[ = \frac{1}{s^2} - \frac{1}{s^2}e^{-s} \] Factor: \[ \mathcal{L}\{f(t)\} = \frac{1 - e^{-s}}{s^2}, \quad s>0. \]
Common Laplace Transforms
| Function \(f(t)\) | \(\mathcal{L}\{f(t)\}\) | Condition |
|---|---|---|
| \(k\) | \(\frac{k}{s}\) | \(s>0\) |
| \(t^n\) | \(\frac{n!}{s^{n+1}}\) | \(s>0\) |
| \(e^{at}\) | \(\frac{1}{s-a}\) | \(s>a\) |
| \(\sin(kt)\) | \(\frac{k}{s^2+k^2}\) | \(s>0\) |
| \(\cos(kt)\) | \(\frac{s}{s^2+k^2}\) | \(s>0\) |
Lecture 17: Section 7.2 - Inverse Transform ↑ Back to top
Inverse Transform
To accompany the Laplace Transform, we have the inverse transform. This allows us to recover the original function that we took the transform of.
To keep the analogy with other transforms, consider:
\[ \frac{d}{dx}(f(x)) = f'(x) \] \[ \int f'(x)\,dx = f(x) + C \]Similarly:
\[ \mathcal{L}^{-1}\{\mathcal{L}\{f(t)\}\} = f(t) \]
Definition.
If \(F(s) = \mathcal{L}\{f(t)\}\), then \[ f(t) = \mathcal{L}^{-1}\{F(s)\} \] is the inverse Laplace transform.
If \(F(s) = \mathcal{L}\{f(t)\}\), then \[ f(t) = \mathcal{L}^{-1}\{F(s)\} \] is the inverse Laplace transform.
The inverse transform is linear:
\[ \mathcal{L}^{-1}\{aF(s) + bG(s)\} = a\mathcal{L}^{-1}\{F(s)\} + b\mathcal{L}^{-1}\{G(s)\} \]
Example 1
Compute \[ \mathcal{L}^{-1}\left\{\frac{6s+1}{s^2+25}\right\} \] Solution: \[ = \mathcal{L}^{-1}\left\{\frac{6s}{s^2+25}\right\} + \mathcal{L}^{-1}\left\{\frac{1}{s^2+25}\right\} \] \[ = 6\mathcal{L}^{-1}\left\{\frac{s}{s^2+25}\right\} + \mathcal{L}^{-1}\left\{\frac{1}{s^2+25}\right\} \] Using known transforms: \[ \mathcal{L}^{-1}\left\{\frac{s}{s^2+a^2}\right\} = \cos(at) \] \[ \mathcal{L}^{-1}\left\{\frac{1}{s^2+a^2}\right\} = \frac{1}{a}\sin(at) \] \[ = 6\cos(5t) + \frac{1}{5}\sin(5t) \]
Compute \[ \mathcal{L}^{-1}\left\{\frac{6s+1}{s^2+25}\right\} \] Solution: \[ = \mathcal{L}^{-1}\left\{\frac{6s}{s^2+25}\right\} + \mathcal{L}^{-1}\left\{\frac{1}{s^2+25}\right\} \] \[ = 6\mathcal{L}^{-1}\left\{\frac{s}{s^2+25}\right\} + \mathcal{L}^{-1}\left\{\frac{1}{s^2+25}\right\} \] Using known transforms: \[ \mathcal{L}^{-1}\left\{\frac{s}{s^2+a^2}\right\} = \cos(at) \] \[ \mathcal{L}^{-1}\left\{\frac{1}{s^2+a^2}\right\} = \frac{1}{a}\sin(at) \] \[ = 6\cos(5t) + \frac{1}{5}\sin(5t) \]
Strategy: Use partial fractions to match known Laplace forms.
Example 2
Find \[ \mathcal{L}^{-1}\left\{\frac{2s}{s^2+3s+2}\right\} \] Solution: Factor: \[ s^2+3s+2 = (s+1)(s+2) \] Partial fractions: \[ \frac{2s}{(s+1)(s+2)} = \frac{A}{s+1} + \frac{B}{s+2} \] Multiply: \[ 2s = A(s+2) + B(s+1) \] Expand: \[ 2s = (A+B)s + (2A+B) \] Match coefficients: \[ A + B = 2, \quad 2A + B = 0 \] Solve: \[ A = -2, \quad B = 4 \] Thus: \[ \frac{2s}{(s+1)(s+2)} = \frac{-2}{s+1} + \frac{4}{s+2} \] Inverse transforms: \[ \mathcal{L}^{-1}\left\{\frac{1}{s-a}\right\} = e^{at} \] \[ = -2e^{-t} + 4e^{-2t} \]
Find \[ \mathcal{L}^{-1}\left\{\frac{2s}{s^2+3s+2}\right\} \] Solution: Factor: \[ s^2+3s+2 = (s+1)(s+2) \] Partial fractions: \[ \frac{2s}{(s+1)(s+2)} = \frac{A}{s+1} + \frac{B}{s+2} \] Multiply: \[ 2s = A(s+2) + B(s+1) \] Expand: \[ 2s = (A+B)s + (2A+B) \] Match coefficients: \[ A + B = 2, \quad 2A + B = 0 \] Solve: \[ A = -2, \quad B = 4 \] Thus: \[ \frac{2s}{(s+1)(s+2)} = \frac{-2}{s+1} + \frac{4}{s+2} \] Inverse transforms: \[ \mathcal{L}^{-1}\left\{\frac{1}{s-a}\right\} = e^{at} \] \[ = -2e^{-t} + 4e^{-2t} \]
Example 3
Find \[ \mathcal{L}^{-1}\left\{\frac{6s-2}{s^2-1}\right\} \] Solution: Factor: \[ s^2 - 1 = (s-1)(s+1) \] Partial fractions: \[ \frac{6s-2}{(s-1)(s+1)} = \frac{A}{s-1} + \frac{B}{s+1} \] Multiply: \[ 6s - 2 = A(s+1) + B(s-1) \] Expand: \[ 6s - 2 = (A+B)s + (A-B) \] Match coefficients: \[ A + B = 6, \quad A - B = -2 \] Solve: \[ A = 2, \quad B = 4 \] Thus: \[ \frac{6s-2}{s^2-1} = \frac{2}{s-1} + \frac{4}{s+1} \] Inverse transforms: \[ = 2e^{t} + 4e^{-t} \]
Find \[ \mathcal{L}^{-1}\left\{\frac{6s-2}{s^2-1}\right\} \] Solution: Factor: \[ s^2 - 1 = (s-1)(s+1) \] Partial fractions: \[ \frac{6s-2}{(s-1)(s+1)} = \frac{A}{s-1} + \frac{B}{s+1} \] Multiply: \[ 6s - 2 = A(s+1) + B(s-1) \] Expand: \[ 6s - 2 = (A+B)s + (A-B) \] Match coefficients: \[ A + B = 6, \quad A - B = -2 \] Solve: \[ A = 2, \quad B = 4 \] Thus: \[ \frac{6s-2}{s^2-1} = \frac{2}{s-1} + \frac{4}{s+1} \] Inverse transforms: \[ = 2e^{t} + 4e^{-t} \]
Example 4
Find \[ \mathcal{L}^{-1}\left\{\frac{s+1}{s^3+25s}\right\} \] Solution: Factor: \[ s^3 + 25s = s(s^2+25) \] Partial fractions: \[ \frac{s+1}{s(s^2+25)} = \frac{A}{s} + \frac{Bs + C}{s^2+25} \] Multiply: \[ s+1 = A(s^2+25) + (Bs+C)s \] Expand: \[ s+1 = (A+B)s^2 + Cs + 25A \] Match coefficients: \[ A + B = 0, \quad C = 1, \quad 25A = 1 \] Solve: \[ A = \frac{1}{25}, \quad B = -\frac{1}{25}, \quad C = 1 \] Thus: \[ = \frac{1}{25}\frac{1}{s} -\frac{1}{25}\frac{s}{s^2+25} + \frac{1}{s^2+25} \] Inverse transforms: \[ \mathcal{L}^{-1}\left\{\frac{1}{s}\right\} = 1 \] \[ \mathcal{L}^{-1}\left\{\frac{s}{s^2+25}\right\} = \cos(5t) \] \[ \mathcal{L}^{-1}\left\{\frac{1}{s^2+25}\right\} = \frac{1}{5}\sin(5t) \] Final answer: \[ = \frac{1}{25} - \frac{1}{25}\cos(5t) + \frac{1}{5}\sin(5t) \]
Find \[ \mathcal{L}^{-1}\left\{\frac{s+1}{s^3+25s}\right\} \] Solution: Factor: \[ s^3 + 25s = s(s^2+25) \] Partial fractions: \[ \frac{s+1}{s(s^2+25)} = \frac{A}{s} + \frac{Bs + C}{s^2+25} \] Multiply: \[ s+1 = A(s^2+25) + (Bs+C)s \] Expand: \[ s+1 = (A+B)s^2 + Cs + 25A \] Match coefficients: \[ A + B = 0, \quad C = 1, \quad 25A = 1 \] Solve: \[ A = \frac{1}{25}, \quad B = -\frac{1}{25}, \quad C = 1 \] Thus: \[ = \frac{1}{25}\frac{1}{s} -\frac{1}{25}\frac{s}{s^2+25} + \frac{1}{s^2+25} \] Inverse transforms: \[ \mathcal{L}^{-1}\left\{\frac{1}{s}\right\} = 1 \] \[ \mathcal{L}^{-1}\left\{\frac{s}{s^2+25}\right\} = \cos(5t) \] \[ \mathcal{L}^{-1}\left\{\frac{1}{s^2+25}\right\} = \frac{1}{5}\sin(5t) \] Final answer: \[ = \frac{1}{25} - \frac{1}{25}\cos(5t) + \frac{1}{5}\sin(5t) \]
Lecture 18: Section 7.2 (cont'd) - Laplace Transforms to Solve IVPs ↑ Back to top
Solving Initial Value Problems using Laplace Transforms
From the last two lectures, we know how to take Laplace transforms and recover functions using inverse transforms. To solve differential equations, we need to understand how Laplace transforms interact with derivatives.
Theorem.
Let \(F(s) = \mathcal{L}\{y(t)\}\). Then \[ \mathcal{L}\{y^{(n)}(t)\} = s^nF(s) - s^{n-1}y(0) - \cdots - sy^{(n-2)}(0) - y^{(n-1)}(0). \]
Let \(F(s) = \mathcal{L}\{y(t)\}\). Then \[ \mathcal{L}\{y^{(n)}(t)\} = s^nF(s) - s^{n-1}y(0) - \cdots - sy^{(n-2)}(0) - y^{(n-1)}(0). \]
This tells us that solving an IVP using Laplace transforms reduces the problem to algebra, provided we know the initial conditions.
Remark.
For first and second derivatives: \[ \mathcal{L}\{y'(t)\} = sF(s) - y(0), \] \[ \mathcal{L}\{y''(t)\} = s^2F(s) - sy(0) - y'(0). \]
For first and second derivatives: \[ \mathcal{L}\{y'(t)\} = sF(s) - y(0), \] \[ \mathcal{L}\{y''(t)\} = s^2F(s) - sy(0) - y'(0). \]
Procedure
- Take the Laplace transform of both sides.
- Solve algebraically for \(F(s)\).
- Take the inverse Laplace transform to find \(y(t)\).
Example 1
Use the Laplace transform to solve \[ y' + 4y = 16e^t, \quad y(0)=1. \] Solution: Take Laplace transforms: \[ sF(s) - 1 + 4F(s) = \frac{16}{s-1}. \] \[ (s+4)F(s) = \frac{16}{s-1} + 1. \] Write \(1 = \dfrac{s-1}{s-1}\): \[ (s+4)F(s) = \frac{16 + (s-1)}{s-1} = \frac{s+15}{s-1}. \] \[ F(s) = \frac{s+15}{(s-1)(s+4)}. \] Partial Fractions: \[ \frac{s+15}{(s-1)(s+4)} = \frac{A}{s-1} + \frac{B}{s+4}. \] \[ s+15 = A(s+4) + B(s-1). \] Expand: \[ s+15 = (A+B)s + (4A - B). \] Match coefficients: \[ A+B = 1, \quad 4A - B = 15. \] Solve: \[ A = \frac{16}{5}, \quad B = -\frac{11}{5}. \] Thus, \[ F(s) = \frac{16/5}{s-1} - \frac{11/5}{s+4}. \] \[ y(t) = \frac{16}{5}e^t - \frac{11}{5}e^{-4t}. \]
Use the Laplace transform to solve \[ y' + 4y = 16e^t, \quad y(0)=1. \] Solution: Take Laplace transforms: \[ sF(s) - 1 + 4F(s) = \frac{16}{s-1}. \] \[ (s+4)F(s) = \frac{16}{s-1} + 1. \] Write \(1 = \dfrac{s-1}{s-1}\): \[ (s+4)F(s) = \frac{16 + (s-1)}{s-1} = \frac{s+15}{s-1}. \] \[ F(s) = \frac{s+15}{(s-1)(s+4)}. \] Partial Fractions: \[ \frac{s+15}{(s-1)(s+4)} = \frac{A}{s-1} + \frac{B}{s+4}. \] \[ s+15 = A(s+4) + B(s-1). \] Expand: \[ s+15 = (A+B)s + (4A - B). \] Match coefficients: \[ A+B = 1, \quad 4A - B = 15. \] Solve: \[ A = \frac{16}{5}, \quad B = -\frac{11}{5}. \] Thus, \[ F(s) = \frac{16/5}{s-1} - \frac{11/5}{s+4}. \] \[ y(t) = \frac{16}{5}e^t - \frac{11}{5}e^{-4t}. \]
Example 2
Use the Laplace transform to solve \[ y'' + 5y' + 4y = 0, \quad y(0)=1,\; y'(0)=0. \] Solution: Take Laplace transforms: \[ (s^2F(s) - s) + 5(sF(s) - 1) + 4F(s) = 0. \] \[ s^2F(s) - s + 5sF(s) - 5 + 4F(s) = 0. \] \[ (s^2 + 5s + 4)F(s) = s + 5. \] Factor: \[ s^2 + 5s + 4 = (s+1)(s+4). \] \[ F(s) = \frac{s+5}{(s+1)(s+4)}. \] Partial Fractions: \[ \frac{s+5}{(s+1)(s+4)} = \frac{A}{s+1} + \frac{B}{s+4}. \] \[ s+5 = A(s+4) + B(s+1). \] \[ s+5 = (A+B)s + (4A + B). \] Match: \[ A+B=1, \quad 4A+B=5. \] Solve: \[ A=\frac{4}{3}, \quad B=-\frac{1}{3}. \] Thus, \[ F(s) = \frac{4/3}{s+1} - \frac{1/3}{s+4}. \] \[ y(t) = \frac{4}{3}e^{-t} - \frac{1}{3}e^{-4t}. \]
Use the Laplace transform to solve \[ y'' + 5y' + 4y = 0, \quad y(0)=1,\; y'(0)=0. \] Solution: Take Laplace transforms: \[ (s^2F(s) - s) + 5(sF(s) - 1) + 4F(s) = 0. \] \[ s^2F(s) - s + 5sF(s) - 5 + 4F(s) = 0. \] \[ (s^2 + 5s + 4)F(s) = s + 5. \] Factor: \[ s^2 + 5s + 4 = (s+1)(s+4). \] \[ F(s) = \frac{s+5}{(s+1)(s+4)}. \] Partial Fractions: \[ \frac{s+5}{(s+1)(s+4)} = \frac{A}{s+1} + \frac{B}{s+4}. \] \[ s+5 = A(s+4) + B(s+1). \] \[ s+5 = (A+B)s + (4A + B). \] Match: \[ A+B=1, \quad 4A+B=5. \] Solve: \[ A=\frac{4}{3}, \quad B=-\frac{1}{3}. \] Thus, \[ F(s) = \frac{4/3}{s+1} - \frac{1/3}{s+4}. \] \[ y(t) = \frac{4}{3}e^{-t} - \frac{1}{3}e^{-4t}. \]
Example 3
Use the Laplace transform to solve \[ y'' + 4y = 10\cos(5t), \quad y(0)=0,\; y'(0)=0. \] Solution: Take Laplace transforms: \[ s^2F(s) + 4F(s) = \frac{10s}{s^2+25}. \] \[ (s^2+4)F(s) = \frac{10s}{s^2+25}. \] \[ F(s) = \frac{10s}{(s^2+4)(s^2+25)}. \] Partial Fractions: \[ \frac{10s}{(s^2+4)(s^2+25)} = \frac{As+B}{s^2+4} + \frac{Cs+D}{s^2+25}. \] Multiply: \[ 10s = (As+B)(s^2+25) + (Cs+D)(s^2+4). \] Expand: \[ 10s = (A+C)s^3 + (B+D)s^2 + (25A+4C)s + (25B+4D). \] Match coefficients: \[ A+C=0, \quad B+D=0, \] \[ 25A+4C=10, \quad 25B+4D=0. \] Solve: \[ C=-A,\quad D=-B. \] \[ 25A - 4A = 10 \Rightarrow 21A=10 \Rightarrow A=\frac{10}{21},\quad C=-\frac{10}{21}. \] \[ 25B - 4B = 0 \Rightarrow 21B=0 \Rightarrow B=0,\quad D=0. \] Thus, \[ F(s) = \frac{10/21 \, s}{s^2+4} - \frac{10/21 \, s}{s^2+25}. \] Take inverse transforms: \[ \mathcal{L}^{-1}\left\{\frac{s}{s^2+4}\right\} = \cos(2t), \quad \mathcal{L}^{-1}\left\{\frac{s}{s^2+25}\right\} = \cos(5t). \] \[ y(t) = \frac{10}{21}\cos(2t) - \frac{10}{21}\cos(5t). \]
Use the Laplace transform to solve \[ y'' + 4y = 10\cos(5t), \quad y(0)=0,\; y'(0)=0. \] Solution: Take Laplace transforms: \[ s^2F(s) + 4F(s) = \frac{10s}{s^2+25}. \] \[ (s^2+4)F(s) = \frac{10s}{s^2+25}. \] \[ F(s) = \frac{10s}{(s^2+4)(s^2+25)}. \] Partial Fractions: \[ \frac{10s}{(s^2+4)(s^2+25)} = \frac{As+B}{s^2+4} + \frac{Cs+D}{s^2+25}. \] Multiply: \[ 10s = (As+B)(s^2+25) + (Cs+D)(s^2+4). \] Expand: \[ 10s = (A+C)s^3 + (B+D)s^2 + (25A+4C)s + (25B+4D). \] Match coefficients: \[ A+C=0, \quad B+D=0, \] \[ 25A+4C=10, \quad 25B+4D=0. \] Solve: \[ C=-A,\quad D=-B. \] \[ 25A - 4A = 10 \Rightarrow 21A=10 \Rightarrow A=\frac{10}{21},\quad C=-\frac{10}{21}. \] \[ 25B - 4B = 0 \Rightarrow 21B=0 \Rightarrow B=0,\quad D=0. \] Thus, \[ F(s) = \frac{10/21 \, s}{s^2+4} - \frac{10/21 \, s}{s^2+25}. \] Take inverse transforms: \[ \mathcal{L}^{-1}\left\{\frac{s}{s^2+4}\right\} = \cos(2t), \quad \mathcal{L}^{-1}\left\{\frac{s}{s^2+25}\right\} = \cos(5t). \] \[ y(t) = \frac{10}{21}\cos(2t) - \frac{10}{21}\cos(5t). \]
Lecture 19: Section 7.3 - Translation Theorems ↑ Back to top
In this lecture, we derive and study the translation theorems. These theorems allow us to use known Laplace transforms \(\mathcal{L}\{f(t)\}\) to compute transforms of modified functions such as \(e^{at}f(t)\) and shifted functions involving unit step functions.
First Translation Theorem
Suppose \(\mathcal{L}\{f(t)\} = F(s)\). Then
\[ \mathcal{L}\{e^{at}f(t)\} = \int_0^\infty e^{-st}e^{at}f(t)\,dt = \int_0^\infty e^{-(s-a)t}f(t)\,dt = F(s-a). \]
Theorem.
If \(F(s) = \mathcal{L}\{f(t)\}\), then \[ \mathcal{L}\{e^{at}f(t)\} = F(s-a). \]
If \(F(s) = \mathcal{L}\{f(t)\}\), then \[ \mathcal{L}\{e^{at}f(t)\} = F(s-a). \]
Example 1
Find \[ \mathcal{L}\{e^{3t}\cos(5t)\}. \] Solution: We know: \[ \mathcal{L}\{\cos(5t)\} = \frac{s}{s^2+25}. \] Apply the shift \(s \to s-3\): \[ \mathcal{L}\{e^{3t}\cos(5t)\} = \frac{s-3}{(s-3)^2+25}. \]
Find \[ \mathcal{L}\{e^{3t}\cos(5t)\}. \] Solution: We know: \[ \mathcal{L}\{\cos(5t)\} = \frac{s}{s^2+25}. \] Apply the shift \(s \to s-3\): \[ \mathcal{L}\{e^{3t}\cos(5t)\} = \frac{s-3}{(s-3)^2+25}. \]
Example 2
Solve \[ y'' + y = te^t,\quad y(0)=1,\; y'(0)=0. \] Step 1: Take Laplace transforms \[ \mathcal{L}\{y''\} = s^2F(s) - sy(0) - y'(0) = s^2F(s)-s. \] We setup the equation \[ (s^2+1)F(s) - s = \mathcal{L}\{te^t\} \] Using translation: \[ \mathcal{L}\{te^t\} = \frac{1}{(s-1)^2} \] \[ (s^2+1)F(s) = s + \frac{1}{(s-1)^2} \] We have \[ F(s) = \frac{s}{s^2+1} + \frac{1}{(s-1)^2(s^2+1)} \] Step 2: Partial fractions \[ \frac{1}{(s-1)^2(s^2+1)} = \frac{A}{s-1} + \frac{B}{(s-1)^2} + \frac{Cs+D}{s^2+1} \] After solving: \[ A=-\frac{1}{2},\quad B=\frac{1}{2},\quad C=\frac{1}{2},\quad D=0 \] \[ = -\frac{1}{2}\frac{1}{s-1} + \frac{1}{2}\frac{1}{(s-1)^2} + \frac{1}{2}\frac{s}{s^2+1} \] Step 3: Inverse Laplace \[ \mathcal{L}^{-1}\left\{\frac{s}{s^2+1}\right\} = \cos t \] \[ \mathcal{L}^{-1}\left\{\frac{1}{s-1}\right\} = e^t \quad \mathcal{L}^{-1}\left\{\frac{1}{(s-1)^2}\right\} = te^t \] Thus: \[ y(t) = \cos t + \left(-\frac{1}{2}e^t + \frac{1}{2}te^t + \frac{1}{2}\cos t\right) \] Final Answer \[ y(t) = \frac{3}{2}\cos t + \frac{1}{2}te^t - \frac{1}{2}e^t. \]
Solve \[ y'' + y = te^t,\quad y(0)=1,\; y'(0)=0. \] Step 1: Take Laplace transforms \[ \mathcal{L}\{y''\} = s^2F(s) - sy(0) - y'(0) = s^2F(s)-s. \] We setup the equation \[ (s^2+1)F(s) - s = \mathcal{L}\{te^t\} \] Using translation: \[ \mathcal{L}\{te^t\} = \frac{1}{(s-1)^2} \] \[ (s^2+1)F(s) = s + \frac{1}{(s-1)^2} \] We have \[ F(s) = \frac{s}{s^2+1} + \frac{1}{(s-1)^2(s^2+1)} \] Step 2: Partial fractions \[ \frac{1}{(s-1)^2(s^2+1)} = \frac{A}{s-1} + \frac{B}{(s-1)^2} + \frac{Cs+D}{s^2+1} \] After solving: \[ A=-\frac{1}{2},\quad B=\frac{1}{2},\quad C=\frac{1}{2},\quad D=0 \] \[ = -\frac{1}{2}\frac{1}{s-1} + \frac{1}{2}\frac{1}{(s-1)^2} + \frac{1}{2}\frac{s}{s^2+1} \] Step 3: Inverse Laplace \[ \mathcal{L}^{-1}\left\{\frac{s}{s^2+1}\right\} = \cos t \] \[ \mathcal{L}^{-1}\left\{\frac{1}{s-1}\right\} = e^t \quad \mathcal{L}^{-1}\left\{\frac{1}{(s-1)^2}\right\} = te^t \] Thus: \[ y(t) = \cos t + \left(-\frac{1}{2}e^t + \frac{1}{2}te^t + \frac{1}{2}\cos t\right) \] Final Answer \[ y(t) = \frac{3}{2}\cos t + \frac{1}{2}te^t - \frac{1}{2}e^t. \]
Second Translation Theorem
Definition.
The unit step function is \[ \mathscr{U}(t-a)= \begin{cases} 0, & 0 \le t < a \\ 1, & t \ge a \end{cases} \]
The unit step function is \[ \mathscr{U}(t-a)= \begin{cases} 0, & 0 \le t < a \\ 1, & t \ge a \end{cases} \]
Theorem (Second Translation).
If \(F(s)=\mathcal{L}\{f(t)\}\), then \[ \mathcal{L}\{f(t-a)\mathscr{U}(t-a)\} = e^{-as}F(s). \]
If \(F(s)=\mathcal{L}\{f(t)\}\), then \[ \mathcal{L}\{f(t-a)\mathscr{U}(t-a)\} = e^{-as}F(s). \]
Proof:
\[ \mathcal{L}\{f(t-a)\mathscr{U}(t-a)\} = \int_0^\infty e^{-st} f(t-a)\mathscr{U}(t-a)\,dt \] \[ = \int_a^\infty e^{-st} f(t-a)\,dt \] Let \(u=t-a\): \[ = \int_0^\infty e^{-s(u+a)} f(u)\,du = e^{-as} \int_0^\infty e^{-su} f(u)\,du = e^{-as}F(s) \]
Remark.
Expressions must be written in the form \(f(t-a)\mathscr{U}(t-a)\).
Expressions must be written in the form \(f(t-a)\mathscr{U}(t-a)\).
Example 3
Write \(f(t)\) in terms of unit step functions, where \[ f(t)= \begin{cases} 1, & 0 \le t \le 3 \\ t^2, & t>3 \end{cases} \] Solution: \[ f(t) = 1 + (t^2-1)\mathscr{U}(t-3) \] Rewrite: \[ t^2 = (t-3)^2 + 6(t-3) + 9 \] \[ t^2 - 1 = (t-3)^2 + 6(t-3) + 8 \] \[ f(t) = 1 + [(t-3)^2 + 6(t-3) + 8]\mathscr{U}(t-3) \]
Write \(f(t)\) in terms of unit step functions, where \[ f(t)= \begin{cases} 1, & 0 \le t \le 3 \\ t^2, & t>3 \end{cases} \] Solution: \[ f(t) = 1 + (t^2-1)\mathscr{U}(t-3) \] Rewrite: \[ t^2 = (t-3)^2 + 6(t-3) + 9 \] \[ t^2 - 1 = (t-3)^2 + 6(t-3) + 8 \] \[ f(t) = 1 + [(t-3)^2 + 6(t-3) + 8]\mathscr{U}(t-3) \]
Example 4
Find \(\mathcal{L}\{f(t)\}\) where \[ f(t)= \begin{cases} 1, & 0 \le t \le 2 \\ t+3, & t>2 \end{cases} \] Solution: \[ f(t)=1+(t+2)\mathscr{U}(t-2) \] Rewrite: \[ t+2=(t-2)+4 \] \[ f(t)=1+[(t-2)+4]\mathscr{U}(t-2) \] Take transforms: \[ \mathcal{L}\{f(t)\} = \frac{1}{s} + e^{-2s}\left(\frac{1}{s^2}+\frac{4}{s}\right) \]
Find \(\mathcal{L}\{f(t)\}\) where \[ f(t)= \begin{cases} 1, & 0 \le t \le 2 \\ t+3, & t>2 \end{cases} \] Solution: \[ f(t)=1+(t+2)\mathscr{U}(t-2) \] Rewrite: \[ t+2=(t-2)+4 \] \[ f(t)=1+[(t-2)+4]\mathscr{U}(t-2) \] Take transforms: \[ \mathcal{L}\{f(t)\} = \frac{1}{s} + e^{-2s}\left(\frac{1}{s^2}+\frac{4}{s}\right) \]
Example 5
Find \[ \mathcal{L}\{\sin(t)\mathscr{U}(t-2\pi)\} \] Solution: \[ \sin(t)=\sin(t-2\pi) \] \[ \mathcal{L}\{\sin(t)\mathscr{U}(t-2\pi)\} = e^{-2\pi s}\mathcal{L}\{\sin t\} = \frac{e^{-2\pi s}}{s^2+1} \]
Find \[ \mathcal{L}\{\sin(t)\mathscr{U}(t-2\pi)\} \] Solution: \[ \sin(t)=\sin(t-2\pi) \] \[ \mathcal{L}\{\sin(t)\mathscr{U}(t-2\pi)\} = e^{-2\pi s}\mathcal{L}\{\sin t\} = \frac{e^{-2\pi s}}{s^2+1} \]
Example 6
Solve \[ y'' + 4y = f(t),\quad y(0)=0,\; y'(0)=0 \] \[ f(t)= \begin{cases} t+1, & 0 \le t \le 1 \\ t-1, & t>1 \end{cases} \] Step 1: Rewrite forcing function using unit step Start with the first expression: \[ f(t) = t+1 \] At \(t=1\), the function changes from \(t+1\) to \(t-1\). So we correct it using a step function: \[ f(t) = (t+1) + [(t-1) - (t+1)]\mathscr{U}(t-1) \] Simplify: \[ (t-1) - (t+1) = -2 \] Thus: \[ f(t) = (t+1) - 2\mathscr{U}(t-1) \] Step 2: Take Laplace transforms Let \(F(s)=\mathcal{L}\{y(t)\}\) \[ \mathcal{L}\{y''\} = s^2F(s) - sy(0) - y'(0) \] Since \(y(0)=0,\; y'(0)=0\): \[ \mathcal{L}\{y''\} = s^2F(s) \] So the equation becomes: \[ (s^2+4)F(s) = \mathcal{L}\{t+1\} - 2\mathcal{L}\{\mathscr{U}(t-1)\} \] Step 3: Compute each Laplace transform \[ \mathcal{L}\{t\} = \frac{1}{s^2}, \quad \mathcal{L}\{1\} = \frac{1}{s} \] So: \[ \mathcal{L}\{t+1\} = \frac{1}{s^2} + \frac{1}{s} \] For the step function: \[ \mathcal{L}\{\mathscr{U}(t-1)\} = \frac{e^{-s}}{s} \] Thus: \[ (s^2+4)F(s) = \frac{1}{s^2} + \frac{1}{s} - \frac{2e^{-s}}{s} \] Step 4: Solve for \(F(s)\) \[ F(s) = \frac{1}{s^2(s^2+4)} + \frac{1}{s(s^2+4)} - \frac{2e^{-s}}{s(s^2+4)} \] Step 5: Inverse Laplace (non-shifted terms) First term: \[ \mathcal{L}^{-1}\left\{\frac{1}{s^2(s^2+4)}\right\} = \frac{t}{4} - \frac{1}{8}\sin(2t) \] Second term: \[ \mathcal{L}^{-1}\left\{\frac{1}{s(s^2+4)}\right\} = \frac{1}{4}(1 - \cos(2t)) \] Step 6: Handle the shifted term We use the Second Translation Theorem: \[ \mathcal{L}^{-1}\left\{\frac{e^{-s}}{s(s^2+4)}\right\} = \left[\mathcal{L}^{-1}\left\{\frac{1}{s(s^2+4)}\right\}\right]_{t \to (t-1)} \mathscr{U}(t-1) \] We already know: \[ \mathcal{L}^{-1}\left\{\frac{1}{s(s^2+4)}\right\} = \frac{1}{4}(1 - \cos(2t)) \] So replace \(t\) with \(t-1\): \[ = \frac{1}{4}\big(1 - \cos(2(t-1))\big)\mathscr{U}(t-1) \] Now multiply by \(-2\): \[ = -\frac{1}{2}\big(1 - \cos(2(t-1))\big)\mathscr{U}(t-1) \] Step 7: Final solution Combine all pieces: \[ y(t) = \left(\frac{t}{4} - \frac{1}{8}\sin(2t)\right) + \frac{1}{4}(1 - \cos(2t)) - \frac{1}{2}(1 - \cos(2(t-1)))\mathscr{U}(t-1) \]
Solve \[ y'' + 4y = f(t),\quad y(0)=0,\; y'(0)=0 \] \[ f(t)= \begin{cases} t+1, & 0 \le t \le 1 \\ t-1, & t>1 \end{cases} \] Step 1: Rewrite forcing function using unit step Start with the first expression: \[ f(t) = t+1 \] At \(t=1\), the function changes from \(t+1\) to \(t-1\). So we correct it using a step function: \[ f(t) = (t+1) + [(t-1) - (t+1)]\mathscr{U}(t-1) \] Simplify: \[ (t-1) - (t+1) = -2 \] Thus: \[ f(t) = (t+1) - 2\mathscr{U}(t-1) \] Step 2: Take Laplace transforms Let \(F(s)=\mathcal{L}\{y(t)\}\) \[ \mathcal{L}\{y''\} = s^2F(s) - sy(0) - y'(0) \] Since \(y(0)=0,\; y'(0)=0\): \[ \mathcal{L}\{y''\} = s^2F(s) \] So the equation becomes: \[ (s^2+4)F(s) = \mathcal{L}\{t+1\} - 2\mathcal{L}\{\mathscr{U}(t-1)\} \] Step 3: Compute each Laplace transform \[ \mathcal{L}\{t\} = \frac{1}{s^2}, \quad \mathcal{L}\{1\} = \frac{1}{s} \] So: \[ \mathcal{L}\{t+1\} = \frac{1}{s^2} + \frac{1}{s} \] For the step function: \[ \mathcal{L}\{\mathscr{U}(t-1)\} = \frac{e^{-s}}{s} \] Thus: \[ (s^2+4)F(s) = \frac{1}{s^2} + \frac{1}{s} - \frac{2e^{-s}}{s} \] Step 4: Solve for \(F(s)\) \[ F(s) = \frac{1}{s^2(s^2+4)} + \frac{1}{s(s^2+4)} - \frac{2e^{-s}}{s(s^2+4)} \] Step 5: Inverse Laplace (non-shifted terms) First term: \[ \mathcal{L}^{-1}\left\{\frac{1}{s^2(s^2+4)}\right\} = \frac{t}{4} - \frac{1}{8}\sin(2t) \] Second term: \[ \mathcal{L}^{-1}\left\{\frac{1}{s(s^2+4)}\right\} = \frac{1}{4}(1 - \cos(2t)) \] Step 6: Handle the shifted term We use the Second Translation Theorem: \[ \mathcal{L}^{-1}\left\{\frac{e^{-s}}{s(s^2+4)}\right\} = \left[\mathcal{L}^{-1}\left\{\frac{1}{s(s^2+4)}\right\}\right]_{t \to (t-1)} \mathscr{U}(t-1) \] We already know: \[ \mathcal{L}^{-1}\left\{\frac{1}{s(s^2+4)}\right\} = \frac{1}{4}(1 - \cos(2t)) \] So replace \(t\) with \(t-1\): \[ = \frac{1}{4}\big(1 - \cos(2(t-1))\big)\mathscr{U}(t-1) \] Now multiply by \(-2\): \[ = -\frac{1}{2}\big(1 - \cos(2(t-1))\big)\mathscr{U}(t-1) \] Step 7: Final solution Combine all pieces: \[ y(t) = \left(\frac{t}{4} - \frac{1}{8}\sin(2t)\right) + \frac{1}{4}(1 - \cos(2t)) - \frac{1}{2}(1 - \cos(2(t-1)))\mathscr{U}(t-1) \]
Lecture 20: Section 7.4 - Derivatives of Transforms and Convolutions ↑ Back to top
In this lecture, we derive and study the relationship between the Laplace transform and other transforms such as the derivative and integral.
Derivative of Transform
Let \( F(s) = \mathcal{L}\{f(t)\} = \int_0^\infty e^{-st} f(t)\,dt \). We want to differentiate \(F(s)\) with respect to \(s\):
\[ F'(s) = \frac{d}{ds} \int_0^\infty e^{-st} f(t)\,dt \]We interchange the derivative and integral. That is,
\[ F'(s) = \int_0^\infty \frac{\partial}{\partial s}(e^{-st}) f(t)\,dt = \int_0^\infty (-t e^{-st}) f(t)\,dt \]Thus,
\[ F'(s) = - \int_0^\infty t f(t) e^{-st}\,dt = -\mathcal{L}\{t f(t)\}. \]It appears that taking the transform \( \mathcal{L}\{t^nf(t)\} \) would be related (up to some sort of potential scaling) to the \(n\)th derivative of \(F(s).\) To get a clean formula, we take the second derivative to get
\[ F''(s) = -\frac{d}{ds} \mathcal{L}\{t f(t)\} = -\left(-\mathcal{L}\{t^2 f(t)\}\right). \]We can see that there is also some alternating component, which leads us to the following theorem:
Theorem
If \( \mathcal{L}\{f(t)\} = F(s) \), then
\[
\mathcal{L}\{t^n f(t)\} = (-1)^n \frac{d^n}{ds^n} F(s)
\]
Example. Find \( \mathcal{L}\{t \sin(4t)\} \).
Solution:
We begin by recalling the Laplace transform:
\[ \mathcal{L}\{\sin(4t)\} = \frac{4}{s^2 + 16}. \]Let \(F(s) = \frac{4}{s^2 + 16}\).
Using the differentiation property of Laplace transforms,
\[ \mathcal{L}\{t f(t)\} = -\frac{d}{ds}F(s), \]we compute the derivative of \(F(s)\).
\[ F'(s) = \frac{d}{ds}\left(\frac{4}{s^2 + 16}\right) = 4 \cdot \frac{d}{ds}(s^2 + 16)^{-1}. \]Applying the chain rule,
\[ F'(s) = 4(-1)(s^2 + 16)^{-2}(2s) = -\frac{8s}{(s^2 + 16)^2}. \]Therefore,
\[ \mathcal{L}\{t \sin(4t)\} = -F'(s) = \frac{8s}{(s^2 + 16)^2}. \]
Example. Find \( \mathcal{L}\{e^{2t} t^2 \cos(t)\} \).
Solution:
We begin by recalling the Laplace transform of \(\cos(t)\):
\[ \mathcal{L}\{\cos(t)\} = \frac{s}{s^2 + 1}. \]Using the first translation theorem,
\[ \mathcal{L}\{e^{at} f(t)\} = F(s-a), \]we replace \(s\) with \(s-2\). Thus,
\[ \mathcal{L}\{e^{2t} \cos(t)\} = \frac{s-2}{(s-2)^2 + 1}. \]Let
\[ F(s) = \frac{s-2}{(s-2)^2 + 1}. \]Step 1: First derivative
Let \(u = s-2\) for simplicity. Then
\[ F(s) = \frac{u}{u^2 + 1}. \]Using the quotient rule,
\[ F'(s) = \frac{(u^2+1)(1) - u(2u)}{(u^2+1)^2} = \frac{u^2 + 1 - 2u^2}{(u^2+1)^2} = \frac{1 - u^2}{(u^2+1)^2}. \]Step 2: Second derivative
\[ F''(s) = \frac{d}{ds}\left(\frac{1 - u^2}{(u^2+1)^2}\right). \] \[ F''(s) = \frac{(-2u)(u^2+1)^2 - (1-u^2)\cdot 2(u^2+1)(2u)}{(u^2+1)^4}. \] \[ F''(s) = \frac{-2u(u^2+1)\left[(u^2+1) + 2(1-u^2)\right]}{(u^2+1)^4}. \] \[ (u^2+1) + 2(1-u^2) = 3 - u^2. \] \[ F''(s) = \frac{-2u(3-u^2)}{(u^2+1)^3}. \]Substitute back \(u = s-2\):
\[ \mathcal{L}\{e^{2t} t^2 \cos(t)\} = F''(s) = \frac{-2(s-2)(3-(s-2)^2)}{\big((s-2)^2+1\big)^3}. \]Convolutions
Definition
Let \(f\) and \(g\) be piecewise continuous functions defined on the interval \([0,\infty)\). Then we defined the convolution of \(f\) and \(g\) as
\[
(f * g)(t) = \int_0^t f(\tau)\, g(t-\tau)\, d\tau
\]
Theorem (Convolution Theorem)
If \( \mathcal{L}\{f(t)\} = F(s) \) and \( \mathcal{L}\{g(t)\} = G(s) \), then
\[ \mathcal{L}\{(f * g)(t)\} = F(s)G(s) \]Remark
This theorem is especially useful for computing inverse Laplace transforms of products. If we encounter a product \(F(s)G(s)\) that is difficult to invert directly, we can instead write:
\[ \mathcal{L}^{-1}\{F(s)G(s)\} = (f * g)(t) \]Example
Find \( \mathcal{L}^{-1}\left\{\frac{1}{s(s-1)}\right\} \) using convolution.
Solution:
We begin by expressing the function as a product:
\[ \frac{1}{s(s-1)} = \frac{1}{s} \cdot \frac{1}{s-1}. \]Recall the inverse Laplace transforms:
\[ \mathcal{L}^{-1}\left\{\frac{1}{s}\right\} = 1, \quad \mathcal{L}^{-1}\left\{\frac{1}{s-1}\right\} = e^{t}. \]By the convolution theorem,
\[ \mathcal{L}^{-1}\left\{\frac{1}{s(s-1)}\right\} = (1 * e^t)(t) = \int_0^t 1 \cdot e^{t-\tau}\, d\tau. \]We now compute the integral:
\[ = \int_0^t e^{t-\tau} d\tau = e^t \int_0^t e^{-\tau} d\tau. \] \[ = e^t \left[ -e^{-\tau} \right]_0^t = e^t (1 - e^{-t}) = e^t - 1. \]Example
Find \( \mathcal{L}^{-1}\left\{\frac{1}{(s+2)(s+3)}\right\} \) using convolution.
Solution:
We rewrite the expression as a product:
\[ \frac{1}{(s+2)(s+3)} = \frac{1}{s+2} \cdot \frac{1}{s+3}. \]Recall:
\[ \mathcal{L}^{-1}\left\{\frac{1}{s+2}\right\} = e^{-2t}, \quad \mathcal{L}^{-1}\left\{\frac{1}{s+3}\right\} = e^{-3t}. \]Using convolution,
\[ (f * g)(t) = \int_0^t e^{-2\tau} e^{-3(t-\tau)}\, d\tau. \]Simplify the integrand:
\[ = \int_0^t e^{-2\tau} e^{-3t + 3\tau} d\tau = e^{-3t} \int_0^t e^{\tau} d\tau. \] \[ = e^{-3t} \left[ e^{\tau} \right]_0^t = e^{-3t}(e^t - 1). \] \[ = e^{-2t} - e^{-3t}. \]Example
Find \( \mathcal{L}^{-1}\left\{\frac{2}{s(s^2+4)}\right\} \) using convolution.
Solution:
We first write the expression as a product:
\[ \frac{2}{s(s^2+4)} = \frac{1}{s} \cdot \frac{2}{s^2+4}. \]Recall:
\[ \mathcal{L}^{-1}\left\{\frac{1}{s}\right\} = 1, \quad \mathcal{L}^{-1}\left\{\frac{2}{s^2+4}\right\} = \sin(2t). \]Thus, using convolution,
\[ \mathcal{L}^{-1}\left\{\frac{2}{s(s^2+4)}\right\} = (1 * \sin(2t))(t) = \int_0^t \sin(2(t-\tau))\, d\tau. \]Let \(u = t - \tau\). Then as \(\tau\) goes from \(0\) to \(t\), \(u\) goes from \(t\) to \(0\). Reversing limits,
\[ = \int_0^t \sin(2u)\, du. \]Now compute the integral:
\[ = \left[-\frac{1}{2}\cos(2u)\right]_0^t = -\frac{1}{2}\cos(2t) + \frac{1}{2}. \] \[ = \frac{1 - \cos(2t)}{2}. \]