Solve by Laplace transforms y"-3y'+2y=2delta(t-1) y(0)=1 y'(0)=0

Harlen Pritchard 2020-11-10 Answered
Solve by Laplace transforms
y"3y+2y=2δ(t1)
y(0)=1
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

Bertha Stark
Answered 2020-11-11 Author has 96 answers
Step 1
The given equation is: 
y"3y+2y=2δ(t1)
Applying the Laplace transform to both sides of the equation, we get 
s2Y(s)sy(0)y(0)3sY(s)y(0)+2Y(s)=2es
Step 2
Substituting the values given in the question, we get 
s2Y(s)sy(0)y(0)3sY(s)+y(0)+2Y(s)=2es
or, Y(s)[s23s+2]s+1=2es
or, Y(s)=2es+s1s23s+2
or, Y(s)=2es+s1s22ss+2
or, Y(s)=2es+s1(s1)(s2)
Step 3
Now, using partial fraction method: 
2es+s1(s1)(s2)=As1+Bs2
or, 2(1s)+s1(s1)(s2)=A(s2)+B(s1)(s1)(s2)
or, 1s(s1)(s2)=s(A+B)(2A+B)(s1)(s2)
Comparing the coefficients:
A+B=1 and 2A+B=1
Solving the above two equations for the value of A and B, we get
A=0 and B=1
Step 4
Substituting the value of A and B in the equation 2es+s1(s1)(s2)=As1+Bs2 we get
2es+s1(s1)(s2)=1s2
taking inverse Laplace transform both sides, we get
L1[2es+s1(s1)(s2)]=L1[1s2]

Expert Community at Your Service

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

New questions