201.ev1.12.Lec14

# S 14 2 47 16 1 2 s 14 s 14 2 47 4 2 5 2 47 47 4 s 14

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s +1/4) 2 + 47 / 16 = 1 2 s +1/4 ( s +1/4) 2 + ( 47 /4 ) 2 5 2 47 47 /4 ( s +1/4) 2 + ( 47 /4 ) 2 . (48) 4 Math 201 Lecture 14: Using Laplace Transform to Solve Equations

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Thus L 1 braceleftbigg s 1 2 s 2 + s +6 bracerightbigg = 1 2 L 1 braceleftBigg s +1/4 ( s +1/4) 2 + ( 47 /4 ) 2 bracerightBigg 5 2 47 L 1 braceleftBigg 47 /4 ( s +1/4) 2 + ( 47 /4 ) 2 bracerightBigg = 1 2 e 1 4 t cos parenleftBigg 47 4 t parenrightBigg 5 2 47 e 1 4 t sin parenleftBigg 47 4 t parenrightBigg . (49) Example of solving equations: Example 8. (7.5 1; 7.5 1) Solve y ′′ 2 y +5 y =0 , y (0)=2 , y (0)=4 (50) using Laplace transform. Solution. We follow the three steps. 1. Transform the equation. a. Transform the LHS. Denote Y = L{ y } . We compute L{ y ′′ } = s 2 Y s y (0) y (0)= s 2 Y 2 s 4 . (51) L{ y } = s Y y (0)= s Y 2 . (52) So L{ y ′′ 2 y +5 y } = s 2 Y 2 s 4 2 sY +8+5 Y =( s 2 2 s +5) Y 2 s. (53) b. Transform the RHS. L{ 0 } =0 . (54) 2. Solve the transformed equation. The transformed equation reads ( s 2 2 s +5) Y 2 s +4=0 Y = 2 s s 2 2 s +5 . (55) 3. Compute y . We have y = L 1 { Y } = 2 L 1 braceleftbigg s ( s 1) 2 +2 2 bracerightbigg = 2 L 1 braceleftBigg s 1 ( s 1) 2 +2 2 + 1 2 2 ( s 1) 2 +2 2 bracerightBigg = 2 e t cos 2 t + e t sin 2 t. (56) When the right hand side is not 0 , taking inverse transform of Y is not as straightfoward as the above anymore. For example, consider the above problem but replace 0 by 1 : y ′′ 2 y +5 y =1 , y (0)=2 , y (0)=4 (57) Transforming the equation we have ( s 2 2 s +5) Y 2 s +4= L{ 1 } = 1 s Y = 2 s s 2 2 s +5 + 1 s ( s 2 2 s +5) . (58) Then Y = L 1 braceleftbigg 2 s s 2 2 s +5 + 1 s ( s 2 2 s +5) bracerightbigg = L 1 braceleftbigg 2 s s 2 2 s +5 bracerightbigg + L 1 braceleftbigg 1 s ( s 2 2 s +5) bracerightbigg . (59) Feb. 8, 2012 5
The question is how to compute the 2nd inverse transform. If we check our inverse transform table, we know how to do inverse transform for A ( s a ) m and A ( s α )+ B β ( s α ) 2 + β 2 , (60) but not for functions like 1 s ( s 2 2 s +5) .
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