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# Answer to Question #231370 in Chemical Engineering for pavani

Question #231370

6) Solve the initial value problem y''-4y=cos 3t ,y(0)=0,y'(0)=1

1
2021-09-29T04:37:09-0400

By Taking Laplace transform to convert t domain into s domain

L(y'') - 4L(y) =L(cos 3t)

s2Y(s)-sy(0)-y'(0)-4Y(s)=(s/s2+32)

Putting y(0)=0 and y'(0)=1 in above equation

s2Y(s)-0-1-4Y(s)= (s/s2+32)

s2Y(s)-1 - 4Y(s)=(s/s2+32)

Y(s)[s2-4]-1= (s/s2+32)

Y(s)[s2-4]= (s/s2+32) + 1

Y(s) = (s/(s2+32)(s2-4)) + 1/(s2-4)

Partial fraction

s/(s2 + 9)(s2-4)= (As+B)/(s2+9) +

Cs+ D/s2-4

s/(s2 + 9)(s2-4)= {(As+B)(s2-4) +

(Cs+ D)(s2+9)}/(s2 + 9)(s2-4)

s=(As+B)(s2-4) + (Cs+ D)(s2+9)

s=As3-4As+Bs2-4B+Cs3+9Cs + Ds2+9D

s=s3(A+C)+s2(B+D)+s(9C-4A)+(9D-4B)

On compairing

A+C=0, B+D=0, 9C-4A=1, 9D-4B=0

A=-C, B=-D, 9C-4(-C)=1

9C+4C=1

13C=1,C=1/13

A=-1/13,. 9D-4B=0,9D-4(-D)=0

9D+4D=0,

13D=0,. D=0, B=-D=0

Y(s)=-(1/13)s/s2+9+(1/13)s/(s2-4) + 1/(s2-4)

Y(s)=-(1/13)s/(s2+9)+(1/13)s/(s2-4) + 1/(s2-4)

Taking Laplace inverse

y(t) = -(1/13)cos3t +(1/13)cosh2t+ 1/4 sinh2t

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