laplace transform

Question

Use laplace transform methods to solve:

 

Y"+2Y+3Y=4e^5t

 

y(0)=-2

Y'(0)=6

Answer 1

To solve this DE by Laplace Transform (LP) we need additivity and the two following properties of LP,

L{y'}=sL{y}-y(0)              <---(1)

L{y''}=s²L{y}-sy(0)-y'(0)  <---(2)

and L{4e^5t}=4(s-5)

 

Given the DE 

y''+2y'+3y=4e^5t

Take LP from both sides

L{y''}+2L{y'}+3L{y}=4/(s-5)

 

Apply (1) and (2):

s²L{y}-sy(0)-y'(0) +2sL{y}-2y(0)+3L{y}=4/(s-5)

Use initial conditions: y(0)=-2, y'(0)=6 and group

(s²+2s+3)L{y}+2s-6+4=4/(s-5)

(s²+2s+3)L{y}=4/(s-5)+2-2s

(s²+2s+3)L{y}=(-2s²+12s-6)/(s-5)

L{y}=(-2s²+12s-6)/[(s-5)(s²+2s+3)] (3)

What we did so far is the LP of the solution. To find the solution, y(t) we hat to take inverse LP from RHS of (3):

 

y(t)=2/19e^-t(e^6t-20cos√2t)+16√2sin(√2))

 

Check all details, there can be mistakes.