Solutions to homework #10

Transcription

1 Solution to homework #0 Problem 7..3 Compute 6 e 3 t t t 8. The firt tep i to ue the linearity of the Laplace tranform to ditribute the tranform over the um and pull the contant factor outide the tranform. That give 6 e 3 t t t 8 6 e 3 t t t 8. Uing the book table, we find e 3 t, 3 (entry 4 with a 3) t, (entry 5 with n ) 3 t, (entry 5 with n ) and Therefore 6 e 3 t t 6 t 8 3 (entry 4 with a 0) Calculation with Mathematica We can alo compute the Laplace tranform automatically uing Mathematica, In[]:= LaplaceTranform 6 Exp 3 t t^ t 8, t, Out[]= which i identical to the reult we got uing table lookup.

2 hwk0soln.nb Problem 7..5 Compute t 3 t e t e 4 t co t. The firt tep i to ue the linearity of the Laplace tranform to ditribute the tranform over the um and pull the contant factor outide the tranform. That give t 3 t e t e 4 t co t t 3 t e t e 4 t co t. Uing the book table, we find 6 t 3, (entry 5 with n 3) 4 t e t, (entry 5 with n and a ) and e 4 t 4 co t 4 Therefore t 3 t e t e 4 t 6 co t 4. (entry 6 with a 4 and b ). 4 4 Calculation with Mathematica We can alo compute the Laplace tranform automatically uing Mathematica, In[3]:= LaplaceTranform t^3 t Exp t Exp 4 t Co t, t, Out[3]= which i identical to the reult we got uing table lookup. Problem Find the invere Laplace tranform of F 0 0 a

3 hwk0soln.nb 3 b^ 4 a c mut be poitive. Computing the dicriminant with a=, b=, and c=0 give the value 36, o the denominator doen t factor in real number. Therefore we proceed to complete the quare. Recall that to complete the quare we write b c in the form A B. Expanding the quare give A A B. We then equate coefficient of each power of (which we can do becaue the power of are linearly independent function) to find the equation and b A c A B. With b and c 0 we find A and B 3. To check a completion of quare, plug A and B into A B and expand it out. You might want to try it by hand; it alo eay to do in Mathematica: Expand 3 0 which i the original polynomial. Therefore we ve completed the quare correctly, and we can rewrite F a. 3 Thi i entry #6 in the table on the book back cover, o we can recognize it a the Laplace tranform of f t e t co 3 t. Calculation in Mathematica I hould alo point out that you can do the whole work in Mathematica uing the InvereLaplaceTranform function: ILT t InvereLaplaceTranform F,, t 3 t 6 t f t FullSimplify ILT t t co 3 t which i identical to the reult found by algebraic manipulation and table lookup. Finally, we can check by computing the Laplace tranform of the anwer LaplaceTranform Exp t Co 3 t, t, 9 Together ExpandAll 0 which i the F with which we tarted. Therefore our calculation check.

4 4 hwk0soln.nb Problem Find the invere Laplace tranform of F The denominator doen t factor in the real (which you can check) o we complete the quare. Uing the procedure for completing the quare hown in the previou problem, the denominator can be rewritten a 4 (which you hould alo check). Looking up F 4 4 Thi i cloe to entry #5 in the table; it differ by a factor of Check by computing the Laplace tranform: LaplaceTranform Exp t Sin t, t, 4 ExpandAll 4 8 which recover the tarting point. Therefore the calculation check. Thu the invere Laplace tranform of F i f t e t in t. Calculation with Mathematica Alternatively, we can ue the InvereLaplaceTranform function to do the problem in one hot: ILT t InvereLaplaceTranform ^ 4 8,, t 4 t 4 t FullSimplify ILT t in t coh t inh t Thi look different, but uing the definition of the hyperbolic function you can how that it i the ame a the reult we got uing table lookup.

5 hwk0soln.nb 5 Problem Table lookup & hand calculation Solve for Y y given y 3 y y co t with initial condition y 0 0 and y 0. Conulting our friendly table, we find y Y y 0 y 0 (entry 4) y Y y 0 (entry 3) o that the Laplace tranform of the LHS i y 3 y y 3 Y. We can alo look up the Laplace tranform of the RHS,. co t Therefore 3 Y from which we can olve for Y, Y. 3 3 Adding the two term give Y. 3 Calculation with Mathematica Here how to do it emi-automatically. Firt compute the Laplace tranform of the LHS In[5]:= Out[5]= LHS LaplaceTranform y t 3 y t y t, t, t y t y 0 t y t 3 t y t y 0 y 0 Becaue Mathematica doen t yet know the initial condition we have to ue tranformation rule to replace y 0 and y 0 with the initial value. Furthermore, the Laplace tranform of the unpecified function y t can t be computed; calling the LaplaceTranform function on it imply return a "placeholder" expreion: In[9]:= LaplaceTranform y t, t, Out[9]= t y t Therefore I ll alo ue a tranformation rule to replace LaplaceTranform[y[t], t, ] with Y. In Mathematica, thi i done a follow: In[6]:= LHS LHS. y 0, y 0 0, LaplaceTranform y t, t, Y reulting in Out[6]= Y 3 Y Y Now I take the Laplace tranform of the RHS In[7]:= Out[7]= RHS LaplaceTranform Co t, t,

6 6 hwk0soln.nb Having computed the Laplace tranform of both ide, we can et up an equation for Y, In[0]:= eqn LHS RHS Out[0]= Y 3 Y Y which we can olve for Y In[8]:= Out[8]= Solve LHS RHS, Y Y 3 Thi i identical to the reult computed by hand.