Fall 2001, AM33 Solution to hw 10 - Applied mathematics

1 fall 2001, am33 Solution to hw 101. Section , problem 9f(t)={t ,if t 2 0,elsewherefis nonzero only between and 2 , so we only need to integrate in that region:L{f}= 2 (t )e stdt= 2 te stdt 2 e stdt=e ss2 e 2 ss2(1 + s)Where the first integral is evaluated by using integration by Section , problem 32f(t)=sint,t [0, )andf(t+ )=f(t). By problem 28L{f}= 0e stsintdt1 e s =1+e s(1 +s2)(1 e s)where the integral is evaluated by applying integration by parts Section , problem 1:y +y=g(t),g(t)={t/2,ift [0,6)3,t 6,y(0) = 0,y (0) = 1 Take the laplace transform of both sides:L{y +y}=L{g}L{y }+L{y}= 60t/2e stdt+ 63e stdtL{y}s2 sy (0) y(0) +L{y}=12s2(1 e 6s)L{y}(s2+1)=12s2(1 e 6s)+1 Thus we have:L{y}=1s2+1+(1 e 6s)1s2(s2+1)L{y}=1s2+1+1s2(s2+1) e 6s12s2(s2+1)1 The partial fraction expansion of1s2(s2+1)is1s2 1s2+1.]]}}

2 Usin gthis , theorem on pa ge 311and the laplace transform table we have that:y(t)=sint+t2 12u6(t)(t 6 sin(t 6))Method 2: Letf(t)= (t)=f(t) u6(t)f(t 6).Therefore,L{g}=(1 e 6s)L{f}=12s2(1 e 6s).The rest is the Section , problem 10 We are given:y +y +5/4y=g(t),y(0) = 0,y (0) = 0,g(t)={sint,ift [0, )0,t We take the laplace transform of both sides:s2L{y} sy(0) y (0) +sL{y} y(0) + 5/4L{y}= 0sinte stdtL{y}(s2+s+5/4) =e s+11+s2L{y}=e s+1(1 +s2)(s2+s+5/4)L{y}=e s+1(1 +s2)((s+1/2)2+1)The partical fraction expansion of1(1+s2)((s+1/2)2+1)is417(4s+3(s+1/2)2+ 1 4s 1s2+1). Usin gthis,theorem , and the laplace transform table one finds:y=h(t)+u (t)h(t )whereh(t)=417( 4cost+sint+4e t/2cost+e t/2sint)5.]}

3 Section , problem 25 We are given:y +2y +2y=f(t);y(0) =y (0) = 0 The homogeneous party +2y +2y= 0 is a constant coefficient problem with chaharacteristicpolynomialr2+2r+2 which have roots: 1 i, 1+i. Therefore, we havey1=e tcost,y2=e tsintas a fundamental set of solutions , and it is easy to check that the WronskianW(y1,y2)(t)=e follows from Section , Problem 22 (page 184), that the Solution to the original IVP isY(t)= t0e scoss e tsint e ssins e tcoste 2sf(s)ds= t0e (t s)f(s)sin(t s)ds(b) Whenf(s)= (s ) the integral becomes:y(t)= t0e (t s) (s )sin(t s)dsFort< , the integration region does not include the pointt= . However, the impulsefunction (t ) put all the mass on pointt= and take value 0 on allt =.

4 Hence theintegral isy(t) = 0 for allt< .However, fort , the integration region includes pointt= , hencey(t)= t0e (t s) (s )sin(t s)ds= e (t s) (s )sin(t s)ds=e (t )sin(t ).It is now easy to see thaty(t)=u (t)e (t )sin(t ).(c) Now we solve the equation with the laplace transform:L{y +2y +2y}=L{ (t )}L{y}s2+L{y}2s+2L{y}=e s L{y}=e s (s+1)2+1By the laplace transform table and the Theorem on page 311 we conclude:y(t)=u (t)e (t )sin(t )6. Section , Problem 3 We are asked to see that sint sintis not positive for everyt. t0sin(t s)sin(s)dsWe can use the trigonometric identity used above: sin(t s)=sintcoss costsins: t0sin(t s)sin(s)ds= t0(sintcoss costsins)sinsds=sint t0cosssinsds cost t0sin2sdsFort=2 the first expression is 0.

5 However the second expression is the integral of apositive function hence it is less than Section Problem 16y +4y +4y=g(t),y(0) = 2,y (0) = 3 LetF(s)=L{y},G(s)=L{g}:L{y +4y +4y}=G(s)s2F(s) sy(0) y (0) + 4sF(s) 4y(0) + 4F(s)=G(s)F(s)(s2+4s+4)=2s+5+G(s)F(s)=2s +5s2+4s+4+G(s)s2+4s+4F(s)=2(s+2)(s+2)2+1 (s+2)2+G(s)(s+2)2F(s)=2s+2+1(s+2)2+G(s)( s+2)2 Usin gthe laplace transform table and Thm (the laplace transform of the convolution isequal to the multiplication of the laplace transforms) on page 331 we have:y(t)=2e 2t+te 2t+ t0(t s)e 2(t s)g(s)ds8. Section Problem 21 Consider the Volterra integral equation: (t)+ t0(t s) (s)ds=sin2twe are givenu (t)= (t). Substitute this in:u (t)+ t0(t s)u (s)ds=sin2tHowever, t0(t s)u (s)ds= t0(t s)d(u (s))=(t s)u (s) t0 t0( 1) u (s)ds= tu (0)+u(t) u(0).

6 It follows thatu (t)+u(t) tu (0) u(0) = sin 2t(b) We are asked the show that the IVP:u (t)+u(t)=sin2t;u(0) = 0,u (0) = 0is equivalent to the integral equation. This means we have to show that given a solutionto the IVP we can find a Solution to the integral equation and vice versa. Assumeu(t)isa Solution to the IVP. Define (t)=u (t). By the calculation above (t) is a Solution to4the integral equation if and only ifu (t)+u(t) tu (0) u(0) = sin conditionsu (0) =u(0) = 0 this reduces tou (t)+u(t)=sin2twhich we know is truebecauseu(t) is a Solution to the IVP assume (t) is a Solution to the integral equation. Defineu(t) such thatu (t)= (t). By the first calculation we knowu(t)satisfies:u (t)+u(t) tu (0) u(0) = sin 2tWe wantu (0) =u(0) = 0 which turns this equation to:u (t)+u(t)=sin2t,u(0) =u (0) = we are we solve the integral equation using the laplace transform:L{ (t)+ t0(t s) (s)ds}=L{sin 2t}F(s)+L{t}F(s)=2s2+4F(s)+1s2F(s)=2s2+4 F(s)s2+1s2=2s2+4F(s)=2s2(s2+4)(s2+1)The partial fraction expansion of the right handside is:23(4s2+4 1s2+1)Usingthelaplacetransform table we find that: (t)=13(4 sin 2t 2sint)The initial value problem can be solved usin gvariation of parameters or method of undeter-mined coefficients it involves nothin Extra questions.

7 By the properties of the convolution(f 1)(t)=(1 f)(t)= t0f(s)dsBy associativity of the convolution ( (f g) h=f (g h)),andthefirstpartwehave:(f 1 1 1) = ((((..((f 1) 1) 1) 1) 1) 1).. 1)= (((..( t0f(tn)dtn) 1) 1) 1).. 1)=((..( t0 tn 10f(tn)dtndtn 1) 1) 1).. 1)..= t0 t10 tn 20 tn 10f(tn)dtndtn 1dtn (t)= t0 t10 tn 20 tn 10f(tn)dtndtn 1dtn take the laplace transform of both sides:L{I(t)}=L{(f 1 1 1)(t)}L{I(t)}=L{f} {I(t)}=L{f}1snL{I(t)}=L{(f tn 1(n 1)!)(t)}which impliesI(t)=(f tn 1(n 1)!)(t)=(tn 1(n 1)! f)(t)=1(n 1)! t0(t s)n 1f(s)ds6