7.2 inverse Laplace Transforms, and application to DEs a lesson for - PowerPoint PPT Presentation

7.2 inverse Laplace Transforms, and application to DEs a lesson for MATH F302 Differential Equations Ed Bueler, Dept. of Mathematics and Statistics, UAF March 22, 2019 for textbook: D. Zill, A First Course in Differential Equations with Modeling Applications , 11th ed. 1 / 18

recall the definition • the Laplace transform of a function f ( t ) defined on (0 , ∞ ) is � ∞ e − st f ( t ) dt L { f ( t ) } = 0 ◦ this is well defined for s > c if f ( t ) has exponential order c : | f ( t ) | ≤ Me ct • the result of applying the Laplace transform is a function of s: L { f ( t ) } = L { f } ( s ) = F ( s ) ← − all mean the same 2 / 18

the Laplace transform strategy old method ODE IVP for y ( t ) y ( t ) = . . . L L − 1 algebraic equation solve for Y Y ( s ) = . . . for Y ( s ) • § 7.2: practice with L − 1 then practice the whole strategy 3 / 18

bring a table to the party • on page 282 of book • this table is pathetic! better one soon . . . 4 / 18

first L − 1 example (like § 7.2 #5) • exercise 1. use algebra and a table of Laplace transforms: � ( s − 1) 3 � L − 1 = s 4 5 / 18

L − 1 example like § 7.2 #11 • exercise 2. use algebra and a table of Laplace transforms: � 5 � L − 1 = s 2 + 36 6 / 18

L − 1 example like § 7.2 #18 • exercise 3. use algebra and a table of Laplace transforms: � s + 1 � L − 1 = s 2 − 7 s 7 / 18

not actually a better table • compare Theorems 7.1.1 and 7.2.1 • they say the same thing! 8 / 18

actually a better table • this substantial table will be printed on your quiz/exam 1 1 1 � e at � � te at � L { 1 } = L = L = ( s − a ) 2 s s − a 1 k n ! � t n e at � L { t } = L { sin( kt ) } = L = s 2 s 2 + k 2 ( s − a ) n +1 n ! s t n � k � L = L { cos( kt ) } = e at sin( kt ) � � s 2 + k 2 L = s n +1 ( s − a ) 2 + k 2 √ π k � t − 1 / 2 � s − a L = L { sinh( kt ) } = e at cos( kt ) � � s 2 − k 2 L = s 1 / 2 ( s − a ) 2 + k 2 √ π s � t 1 / 2 � L { cosh( kt ) } = 2 ks L = s 2 − k 2 2 s 3 / 2 L { t sin( kt ) } = ( s 2 + k 2 ) 2 Γ( α + 1) t α � L � = s 2 − k 2 s α +1 L { t cos( kt ) } = ( s 2 + k 2 ) 2 = ( − 1) s d n � e at f ( t ) � L = F ( s − a ) t n f ( t ) � � L ds n F ( s ) e − as �� t � L {U ( t − a ) } = L f ( τ ) g ( t − τ ) d τ = F ( s ) G ( s ) s 0 L { f ( t − a ) U ( t − a ) } = e − as F ( s ) L { δ ( t ) } = 1 � f ( n ) ( t ) � = s n F ( s ) − s n − 1 f (0) − · · · − f ( n − 1) (0) L { δ ( t − t 0 ) } = e − st 0 L 9 / 18

L − 1 example like § 7.2 #23 • exercise 4. use algebra and a table of Laplace transforms: � � s L − 1 = ( s − 3)( s − 4)( s − 6) 1 2 1 s ( s − 3)( s − 4)( s − 6) = s − 3 − s − 4 + s − 6 10 / 18

L − 1 example like § 7.2 #25 • exercise 5. use algebra and a table of Laplace transforms: � 1 � L − 1 = s 3 + 7 s 11 / 18

transform of first derivatives • exercise 6. suppose F ( s ) = L { f ( t ) } . use the definition of the Laplace transform to show: L { f ′ ( t ) } = s F ( s ) − f (0) • actually we showed this on § 7.1 slides • what assumptions did we make about f ( t )? 12 / 18

transform of second derivatives • exercise 7. suppose F ( s ) = L { f ( t ) } . show: = s 2 F ( s ) − s f (0) − f ′ (0) � � L f ′′ ( t ) • in the table you’ll have in hand during quizzes/exams: � � f ( n ) ( t ) = s n F ( s ) − s n − 1 f (0) − · · · − f ( n − 1) (0) L 13 / 18

like § 7.2 #39 • exercise 8. use Laplace transform to solve the ODE IVP: y ′′ − 5 y ′ + 4 y = 0 , y (0) = 1 , y ′ (0) = 0 14 / 18

the old way • exercise 9. solve without Laplace transform: y ′′ − 5 y ′ + 4 y = 0 , y (0) = 1 , y ′ (0) = 0 15 / 18

like § 7.2 #41 • exercise 10. use Laplace transform to solve the ODE IVP: √ √ y ′′ + y = y (0) = 0 , y ′ (0) = 3 2 cos( 2 t ) , 16 / 18

like § 7.2 #41, cont. √ √ √ y ( t ) = 3 sin( t ) + 2 cos( t ) − 2 cos( 2 t ) 17 / 18

expectations • just watching this video is not enough! ◦ see “found online” videos and stuff at bueler.github.io/math302/week11.html ◦ read section 7.2 (and 7.1 and 7.3) in the textbook ◦ do the WebAssign exercises for section 7.2 18 / 18