Answer to Question #231370 in Chemical Engineering for pavani

Question

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

Answer 1

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