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Someone set us up the bomb (again)
Find solution of initial value problems y″+y=u3π(t), where y(0)=0 and y′(0)=1
ℒ{y″}+ℒ{y}=ℒ{1−u3π(t)}s2ℒ{y}−y′(0)−sy(0)+ℒ{y}=1−e−3πss(s2+1)ℒ{y}−1=1−e−3πssℒ{y}=1s(s2+1)−e−3πss(s2+1)=ℒ{1−cost+sint}−e−3πsℒ{h(t)}=ℒ{1−cost+sint}−ℒ{u3π(t)(1−cos(t−3π))}y=1−cost+sint−u3π(t)(1+cost)={1−cost+sintt<3πsint−2costt≥3π
y″+3y′+2y=u2(t)y(0)=0y′(0)=1
ℒ{y″}+3ℒ{y′}+2ℒ{y}=ℒ{1⋅u2(t)}s2ℒ{y}−1+3sℒ{y}−0+2ℒ{y}=e−2ss(s2+3s+2)ℒ{y}=1+e−2ssℒ{y}=1(s+2)(s+1)+e−2ss(s+2)(s+1)=ℒ{e−t−e−2t}+e−2s(12ℒ{1}+12ℒ{e−2t}−ℒ{e−t})y=e−t−e−2t+u2(t)(12+e−2t2−e−t+2)
h(t)=y″+2y′+2yy(0)=0y′(0)=1h(t)={00<t<π1π≤t<2π02π<t
Rewrite h(t)=0+uπ(t)(1−0)+u2π(t)(0−1)=uπ(t)−u2π(t)
ℒ{y″}+2ℒ{y′}+2ℒ{y}=ℒ{uπ(t)−u2pi(t)}s2ℒ{y}−1+2sℒ{y}−0+2ℒ{y}=e−πss−e−2πss(s2+2s+2)ℒ{y}=1+e−πss−e−2πssℒ{y}=1(s2+1)2+1+(e−πs−e−2πs)(12s−12(s+2(s+1)2+1))=ℒ{e−tsint}+(e−πs−e−2πs)(ℒ{12−12e−tcost+e−tsint})=ℒ{e−tsint}+ℒ{uπ(t)(12−12(e−(t−π)cos(t−π)+e−(t−π)sin(t−π)))}−ℒ{u2π(t)(12−12(e−(t−2π)cos(t−2π)+e−(t−2π)sin(t−2π)))}y=e−tsint+uπ(t)(12−12(e−(t−π)cos(t−π)+e−(t−π)sin(t−π)))−u2π(t)(12−12(e−(t−2π)cos(t−2π)+e−(t−2π)sin(t−2π)))