Transcription of MATH 214 { QUIZ 12 { SOLUTIONS (0) = 1; y (0) = 0
1 math 214 quiz 12 SOLUTIONSUse the Laplace transform (and the table below) to solve the initial value problemy + 3y + 2y= 0,y(0) = 1,y (0) = : Taking the Laplace transform of both sides givesL{y + 3y + 2y}= 0L{y }+ 3L{y }+ 2L{y}= 0s2L{y} s y(0) y (0) + 3(sL{y} y(0)) + 2L{y}= 0(s2+ 3s+ 2)L{y} s 3 = 0so thatL{y}=s+ 3s2+ 3s+ this last term in partial fractions givess+ 3s2+ 3s+ 2=s+ 3(s+ 2)(s+ 1)=As+ 2+Bs+ 1=A(s+ 1) +B(s+ 2)(s+ 2)(s+ 1)so thatA(s+ 1) +B(s+ 2) =s+ ins= 2 givesA= 1 and plugging ins= 1 givesB= 2. Thereforey=L 1{2s+ 1 1s+ 2}= 2e t e the Laplace transform (and the table below) to solve the initial value problemy 4y + 4y= 0,y(0) = 1,y (0) = : Taking the Laplace transform of both sides givesL{y 4y + 4y}= 0L{y } 4L{y }+ 4L{y}= 0s2L{y} s y(0) y (0) 4(sL{y} y(0)) + 4L{y}= 0(s2 4s+ 4)L{y} s+ 3 = 0so thatL{y}=s 3s2 4s+ this last term in partial fractions givess 3s2 4s+ 4=s 3(s 2)2=As 2+B(s 2)2=A(s 2) +B(s 2)2so thatA(s 2) +B=s ins= 2 givesB= 1 and taking a derivative gives immediatelyA= 1.
2 Thereforey=L 1{1s 2 1(s 2)2}=e2t t the Laplace transform (and the table below) to solve the initial value problemy y 6y= 0,y(0) = 1,y (0) = : Taking the Laplace transform of both sides givesL{y y 6y}= 0L{y } L{y } 6L{y}= 0s2L{y} s y(0) y (0) (sL{y} y(0)) 6L{y}= 0(s2 s 6)L{y} s= 0so thatL{y}=ss2 s this last term in partial fractions givess 5s2 s 6=s(s+ 2)(s 3))=As+ 2+B(s 3)=A(s 3) +B(s+ 2)so thatA(s 3) +B(s+ 2) = ins= 2 givesA= 2/5 and plugging ins= 3 givesB= 3/5. Thereforey=L 1{251s+ 2+351(s 3)}=25e 2t+35e3t.
