Find the inverse Laplace transform of the function F (p) = Evaluating the residues at the four simple poles, we find. residue at z = 1 is 4te t

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1 Homework Solutios. Chater, Sectio 7, Problem 56. Fid the iverse Lalace trasform of the fuctio F () (7.6). À Chater, Sectio 7, Problem 6. Fid the iverse Lalace trasform of the fuctio F () usig (7.6). Solutio: by usig ( ) (+) We first fid the oles of F (z)e zt ad factor the deomiator to get F (z)e zt e zt (z )(z+)(z i)(z+). Evaluatig the residues at the four simle oles, we fid The by (7.6) we have residue at z is e t residue at z is e t residue at z i is e it i residue at z i is e it i by f(t) et e t e it i + e it i sih t si t. À We first fid the oles of F (z)e zt ad factor the deomiator to get F (z)e zt (z ) e zt z(z+). Evaluatig the residues at the two oles, we fid The by (7.6) we have residue at z is residue at z is te t f(t) te t.

Evaluatig the residues at the four simle oles, we fid The by (7.

2 . Chater 8, Sectio, Problem 5. Solve the di eretial equatio y + y f(t), y y,withf(t) give by the fuctios i Figure., by the followig methods. (a) Use the covolutio itegral, beig careful to cosider searately the three itervals to t, t to t +,adt + to. (b) Write f (t) asadi ereceofuitstefuctiosasil5, ad use L5 to fid L(f ). Exad ( + ) by artial fractios ad use L8 to fid y (t). Your result should agree with (a). (c) Let ad show that your solutio i (a) ad (b) teds to the same solutio (.5) obtaied usig the fuctios of Figure.3; that is, either set of fuctios gives, i the limit, the same solutio (.) obtaied by usig the fuctio. Note that, whe you let, you do ot eed to cosider the iterval t to t + sice, if t>t, the for su cietly large, t>t +. À Chater 8, Sectio, Problem 7. Usig the fuctio method, fid the resose of y +y + y (t t ) to a uit imulse. Ã Chater 8, Sectio, Problem. Usig the fuctio method, fid the resose of d y dt y (t t )to auitimulse. Solutio: (a) Takig the Lalace trasform of each term, substitutig the iitial coditios, ad solvig for Y,weget Y + Y L(f (t)) Y + L(f (t)) Accordig to L3, the iverse trasform of + Y L si (t) L(f (t)). is si (t).therefore,

$Exad ( + ) by artial fractios ad use L8 to fid y (t). Your result should agree with (a). (c) Let ad show that your solutio i (a) ad (b) teds to the same solutio (.$

3 Accordig to L3, si [(t )] y(t) f ( ) d 8 >< if t<t ( cos [(t t >: )]) if t <t<t + cos [(t t )] cos [(t t )] if t + <t (b) We write f (t) asadi ereceofuitstefuctios: f (t) u(t t ) u(t t ) Accordig to L5, We exad ( + ) L(f (t)) e t e (t + ) by artial fractios to get ( + ) Puttig it all together, we have Y + L(f (t)) + e t e (t+ ) ( + ) e t + e t e ( t + ) e t e ( t + ) + e ( t + ) + e t + Accordig to L8, the iverse Lalace trasform of + e + e ( t + ) e t is u (t t ) e ( t + ) + e t t + is cos ( (t t ))u (t t ) is u t t iscos ( t t )u t t Therefore, 8 >< if t<t y (t) ( cos [(t t >: )]) if t <t<t + cos [(t t )] cos [(t t )] if t + <t 3

Y + L(f (t)) + e t e (t+ ) ( + ) e t + e t e ( t + ) e t e ( t + ) + e ( t + ) + e t + Accordig to L8, the iverse Lalace trasform of + e + e ( t + ) e t is u

4 (c) Lettig,weseethat lim y (t) ( if t<t si ((t t )) if t>t À From Problem 6c, we get the iitial coditios y y. So,takig Lalace trasforms, we get ( +) Y L ( (t t )). Accordig to L6, the Lalace trasform of te t is (+).Therefore, Accordig to L8, Y L te t L ( (t t )). y(t) (t t )e (t t ) u(t t ). Ã Accordig to 35, takig Lalace trasforms yields ( )Y L ( (t t )). Accordig to Problem of HW, the Lalace trasform of sih t si t is.therefore, sih t si t Y L L ( (t t )). Accordig to L8, y(t) sih (t t ) si (t t ) u(t t ).

Therefore, Accordig to L8, Y L te t L ( (t t )). y(t) (t t )e (t t ) u(t t ).

5 3. Chater 8, Sectio, Problem. Use (.6) to solve (.) whe f(t) sit. À Chater 8, Sectio, Problem 6. ( For Problem.7, show that G(t, t <t<t ) sih a(t t ). a t>t Thus write the solutio to.7 as a itegral similar to (.6) ad evaluate it. Ã Chater 8, Sectio, Problem 8. Solve the di eretial equatio y +y + y f(t), y y,where ( <t<a f(t) t > a. As i Problems 6 ad 7, fid the Gree fuctio for the roblem ad use it i y(t) Z Cosider the cases t<aad t>asearately. Solutio: We have from (.6) that y(t) si (t )f( ) d si (t (si t cos G(t, t )f(t ) dt. (.) )si d (si t cos si cos t si t cos t[ t si cos t)si d si cos t si ) d si t ]. 5

Solve the di eretial equatio y +y + y f(t), y y,where ( <t<a f(t) t > a.

6 À The Gree fuctio satisfies d dt G(t, t ) a G(t, t ) (t t ) with G ad d dtg equal to o t. TosolveforG, wetakethelalace trasform of the di eretial equatio to get sih at Accordig to L9, L[ By L8, L[G] Now we solve for y y(t): ( a )L[G] L[ (t t )]. a ] a.therefore e t sih at a L[ a ]L[ (t t )]. G(t, t ) sih a(t t ) a u(t t ). y(t) R G(t, t )f(t ) dt R G(t, t ) dt sih a(t t ) a dt cosh a(t t ) a cosh at a. t t 6

L8, L[G] Now we solve for y y(t): ( a )L[G] L[ (t t )]. a ] a.therefore e t sih at a L[ a ]L[ (t t )].

7 Ã The Gree fuctio satisfies d dt G(t, t )+ d dt G(t, t )+G(t, t ) (t t ) with G ad d dtg equal to o t. TosolveforG, wetakethelalace trasform of the di eretial equatio to get ( + +)L[G] L[ (t t )]. Sice ++ (+), L[te t ] (+).Therefore By L8, L[G] e t (+) L[te t ]L[ (t t )]. G(t, t )(t t )e (t t) u(t t ). Now we solve for y y(t): y(t) R G(t, t )f(t ) dt R a G(t, t ) dt 8R t < (t t )e (t t) dt if <t<a : R a (t t )e (t t) dt if <a<t. Sice R (t t )e (t t) dt (t t +)e (t t), if <t<a,the y(t) (t t +)e (t t ) t t (t +)e t ; if t>a,the y(t) (t t +)e (t t ) a (t a +)e (t a) t (t +)e t. 7

Therefore By L8, L[G] e t (+) L[te t ]L[ (t t )]. G(t, t )(t t )e (t t) u(t t ).

8 . Use y(x) cos x Z x to fid the solutio of (si x)f( x) d x si x Z / (cos x)f( x) d x (.7) y + y f(x) (.7) with y() y(/) whe the forcig fuctio is give f(x). Chater 8, Sectio, Problem. f(x) six À Chater( 8, Sectio, Problem 3. x <x</ f(x) / x / <x</ Solutio: y(x) Z x cos x si x si xd x cos x si x cos x si x si 3 x 6 si 3x 6 x x Z / si x cos x si xd x si x À We have two cases: x</adx>/. For x</, y(x) cos x Z x x si xd x si x Z / si x cos x x cos xd x + cos 3 x 6 Z / / / x ( x)cos xd x cos x (si x x cos x) x x si x (cos x + x si x) / x + si x / cos x (si x x cos x) ( + ) + ( ) ( cos x (si x x cos x) ( ) si x x/ (cos x + x si x) / ) si x x/ 8

x <x</ f(x) / x / <x</ Solutio: y(x) Z x cos x si x si xd x cos x si x cos x si x si 3 x 6 si 3x 6 x x Z / si x cos x si xd x si x À We have two cases: x</adx>/.

9 For x>/, y(x) cos x si x Z / Z / x si xd x + x cos xd x + Z x cos x (si x x cos x) / x cos x ( ) si x ( x)si xd x / Z / / ( x)cos xd x cos x x x/ (si x (cos x si ) x x cos x ( ) si x cos x cos x si x + x cos x ( ) si x x cos x) x x/ 9

Partial Di erential Equations

Partial Di erential Equations Partial Di eretial Equatios Partial Di eretial Equatios Much of moder sciece, egieerig, ad mathematics is based o the study of partial di eretial equatios, where a partial di eretial equatio is a equatio