30.1. LINEAR O.D.E. WITH CONSTANT COEFFICIENTS 583

Table of Laplace Transforms

f (t) F (s)

1.) 1 1/s

2.) eat 1/(s−a)

3.) tn n!sn+1

4.) t p, p >−1 Γ(p+1)sp+1

5.) sinat as2+a2

6.) cosat ss2+a2

7.) eibt s+ibs2+b2

8.) sinhat as2−a2

9.) coshat ss2−a2

10.) eat sinbt b(s−a)2+b2

11.) eat cosbt s−a(s−a)2+b2

f (t) F (s)

12.) eat sinhbt b(s−a)2−b2

13.) eat coshbt s−a(s−a)2−b2

14) tneat n!(s−a)n+1

15.) uc (t) e−cs

s

16.) uc (t) f (t− c) e−csF (s)

17.) ect f (t) F (s− c)

18.) f (ct) 1c F( s

c

)19.) f ∗g(t) F (s)G(s)

20.) δ (t− c) e−cs

21.) f′(t) sF (s)− f (0)

22.) (−t)n f (t) dnFdsn (s)

You should verify the claims in this table. It is best if you do it yourself. The fun-damental result in using Laplace transforms is this. If you have F (s) = G(s) then asidefrom finitely many jumps on each bounded interval, it follows that f (t) = g(t) . Thus youjust go backwards in the table to find the desired functions. To see this shown, see Section4.2 on Page 47. I will illustrate with a second order differential equation having constantcoefficients. Of course you can change to a first order system and this will be the emphasisnext, but you can also use the method directly. Note∫

∞

0y′′ (t)e−stdt = y′ (t)e−st |∞0 + s

∫∞

0y′ (t)e−stdt

= −y′ (0)+ s∫

∞

0y′ (t)e−stdt

= −y′ (0)+ s[

y(t)e−st |∞0 + s∫

∞

0y(t)e−stdt

]= −y′ (0)− sy(0)+ s2Y (s) (30.1)

A similar formula holds for higher derivatives. You can also get this by iterating 21.

Example 30.1.3 Find all solutions to the equation y′′−2y′+ y = e−t .

From the table, first go to y′′. This gives −y′ (0)− sy(0)+ s2Y (s) then you go to thenext term which gives −2sY (s)+2y(0) and finally, you get Y (s) from the y. On the rightyou get from formula 2. 1/(s+1) . Therefore, you have

s2Y (s)−2sY (s)+Y (s)− y′ (0)− sy(0)+2y(0) =1

s+1(s2−2s+1

)Y (s) = y′ (0)+(s−2)y(0)+

1s+1