Solve the question with laplace transformation y''+2y'+y=4 y(0)=3 y'(0)=0

he298c 2021-02-08 Answered
Solve the question with laplace transformation
y+2y+y=4
y(0)=3
y(0)=0
You can still ask an expert for help

Want to know more about Laplace transform?

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

Corben Pittman
Answered 2021-02-09 Author has 83 answers
To solve the differential equation with Laplace Transformation
y+2y+y=4
y(0)=3
y(0)=0
Step 2
Laplace transformation of y'' is
L{y}=s2Ysy(0)y(0)
L{y}=s2Y3s
Laplace transformation of y' is
L{y}=sYy(0)
L{y}=sY3
Laplace transformation of y is
L{y}=Y
Laplace transformation of 4 is
L{4}=4s
Step 3
Laplace transformation of given differential equation
L{y+2y+y}=L{4}
s2Y3s+2(sY3)+Y=4s
s2Y+2sY+Y3s6=4s
(s2+2s+1)Y=4s+3s+6
(s+1)2Y=(3s2+6s+4)s
Y=(3s2+6s+3+1)s(s+1)2
Y=3(s2+2s+1)s(s+1)2+1s(s+1)2
Y=3(s+1)2s(s+1)2+1s(s+1)2
Y=3s+1s(s+1)2
Step 4
1s(s+1)2=As+B(s+1)+C(s+1)2
1=A(s+1)2+Bs(s+1)+(C)s
1=As2+2As+A+Bs2+Bs+Cs
1=(A+B)s2+(2A+B+C)s+A
A=1
A+B=0
1+B=0
B=1
2A+B+C=0
21+C=0
C=1
Thus , 1s(s+1)2=1s+1(s+1)1(s+1)2
Step 5
Thus, from equation (1)
Y=3s+1s1(s+1)1(s+1)2
Y=4s1(s+1)1(s+1)2
Step 6
Inverse Laplace transform of Y is
L1{Y}=y
Inverse Laplace transform of 1s is
L1{1s}=1
Inverse Laplace transform of 1(s+1) is

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more