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Laplace transforms convert functions of time into functions of a complex variable, making many calculus and differential equation problems easier to solve. This cheat sheet helps students recognize standard transforms, apply transform properties, and solve initial value problems efficiently. It is especially useful when working with exponential, trigonometric, polynomial, and piecewise-defined functions.

Key Facts

  • The Laplace transform is defined by L{f(t)}=F(s)=0estf(t)dt\mathcal{L}\{f(t)\}=F(s)=\int_{0}^{\infty} e^{-st}f(t)\,dt when the integral converges.
  • Linearity means L{af(t)+bg(t)}=aF(s)+bG(s)\mathcal{L}\{af(t)+bg(t)\}=aF(s)+bG(s) for constants aa and bb.
  • The basic power rule is L{tn}=n!sn+1\mathcal{L}\{t^n\}=\frac{n!}{s^{n+1}} for integers n0n\ge 0.
  • The exponential shift rule is L{eatf(t)}=F(sa)\mathcal{L}\{e^{at}f(t)\}=F(s-a).
  • Derivative transforms are L{f(t)}=sF(s)f(0)\mathcal{L}\{f'(t)\}=sF(s)-f(0) and L{f(t)}=s2F(s)sf(0)f(0)\mathcal{L}\{f''(t)\}=s^2F(s)-sf(0)-f'(0).
  • The first shifting theorem for step functions is L{u(ta)f(ta)}=easF(s)\mathcal{L}\{u(t-a)f(t-a)\}=e^{-as}F(s) for a>0a>0.
  • Convolution satisfies L{(fg)(t)}=F(s)G(s)\mathcal{L}\{(f*g)(t)\}=F(s)G(s), where (fg)(t)=0tf(τ)g(tτ)dτ(f*g)(t)=\int_0^t f(\tau)g(t-\tau)\,d\tau.
  • Common transforms include L{1}=1s\mathcal{L}\{1\}=\frac{1}{s}, L{eat}=1sa\mathcal{L}\{e^{at}\}=\frac{1}{s-a}, L{sinbt}=bs2+b2\mathcal{L}\{\sin bt\}=\frac{b}{s^2+b^2}, and L{cosbt}=ss2+b2\mathcal{L}\{\cos bt\}=\frac{s}{s^2+b^2}.

Vocabulary

Laplace transform
An integral transform that changes a time-domain function f(t)f(t) into a frequency-domain function F(s)F(s).
Inverse Laplace transform
The operation L1{F(s)}=f(t)\mathcal{L}^{-1}\{F(s)\}=f(t) that recovers the original time-domain function from its transform.
Unit step function
The function u(ta)u(t-a) equals 00 for t<at<a and 11 for tat\ge a, so it models a delayed switch.
Convolution
The convolution (fg)(t)=0tf(τ)g(tτ)dτ(f*g)(t)=\int_0^t f(\tau)g(t-\tau)\,d\tau combines two functions in a way that becomes multiplication after transforming.
Initial value problem
A differential equation together with starting values such as y(0)y(0) and y(0)y'(0).
Region of convergence
The set of ss-values for which the improper integral defining L{f(t)}\mathcal{L}\{f(t)\} converges.

Common Mistakes to Avoid

  • Forgetting initial conditions in derivative transforms is wrong because L{y(t)}=sY(s)y(0)\mathcal{L}\{y'(t)\}=sY(s)-y(0) and L{y(t)}=s2Y(s)sy(0)y(0)\mathcal{L}\{y''(t)\}=s^2Y(s)-sy(0)-y'(0) include starting values.
  • Using F(s+a)F(s+a) instead of F(sa)F(s-a) for eatf(t)e^{at}f(t) is wrong because multiplication by eate^{at} shifts the transform to F(sa)F(s-a).
  • Dropping the factor ease^{-as} with unit step functions is wrong because a delay by aa in time produces the multiplier ease^{-as} in the ss-domain.
  • Taking inverse transforms before partial fraction decomposition is often wrong because expressions such as 2s+3(s1)(s+2)\frac{2s+3}{(s-1)(s+2)} must usually be split into recognizable table forms.
  • Confusing multiplication with convolution is wrong because L1{F(s)G(s)}\mathcal{L}^{-1}\{F(s)G(s)\} is generally (fg)(t)(f*g)(t), not f(t)g(t)f(t)g(t).

Practice Questions

  1. 1 Find L{3t24e5t+2sin(3t)}\mathcal{L}\{3t^2-4e^{5t}+2\sin(3t)\}.
  2. 2 Find L1{6s4+ss2+16}\mathcal{L}^{-1}\{\frac{6}{s^4}+\frac{s}{s^2+16}\}.
  3. 3 Use Laplace transforms to solve y+4y=0y''+4y=0 with y(0)=2y(0)=2 and y(0)=3y'(0)=3.
  4. 4 Explain why Laplace transforms are useful for solving differential equations with discontinuous forcing terms such as u(ta)u(t-a).