Solve the inital value problem by using Laplace transform: y''-5y'+6y=-8cos(t)-2sin(t), y(frac{pi}{2})=1 ,y'(frac{pi}{2})=0

Amari Flowers 2020-11-09 Answered
Solve the inital value problem by using Laplace transform:
y5y+6y=8cos(t)2sin(t),y(π2)=1,y(π2)=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

SchulzD
Answered 2020-11-10 Author has 83 answers
Solve the inital value problem by using Laplace transform
y5y+6y=8cos(t)2sin(t),y(π2)=1,y(π2)=0
Step 2
Consider:y5y+6y=8cos(t)2sin(t),y(π2)=1,y(π2)=0First noticed that the initial condition are not at t=0.We can applythe Laplace transform of derivative is to have the initial condition at t=0.
This mean that we will need to formulate the IVP in such a way thatthe initial condition are at t=0.
We need to change the variable t=η+π2η=tπ2
y(η+π2)5y(η+π2)+6y(η+π2)=8cos(η+π2)2sin(η+π2)
y(η+π2)5y(η+π2)+6y(η+π2)=8sin(η)2cos(η)(2)
We have, u(η)=y(η+π2)
By using chain rule,
u(η)=dudη=dydηdηdt=y(η+π2)
u(η)=dudη=dydηdηdt=y(η+π2)
Hence,the initial condition for u(η) as
u(0)=y(π2)=1 and u(0)=y(π2)=0
Step 3
Now,the equation (2) becomes
u(η)5u(η)+6u(η)=8sin(η)2cos(η)
Apply the Laplace transform on both sides,then
L{u(η)}5L{u(η)}+6L{u(η)}=8L{sin(η)}2L{cos(η)}
s2u(s)su(0)u(0)5su(s)u(0)+6u(s)=8(s2+1)2s(s2+1)
s2u(s)s5su(s)+5+6u(s)=8(s2+1)2s(s2+1)
(s2+1)u(s)=8(s2+1)2s(s2+1)+s5
u(s)=8(s2+1)22s(s2+1)2+s(s2+1
Not exactly what you’re looking for?
Ask My Question

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