
Bobosharif S. answered 06/24/18
Tutor
4.4
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Mathematics/Statistics Tutor
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{4e5t}=4(s-5)
Given the DE
y''+2y'+3y=4e5t
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(e6t-20cos√2t)+16√2sin(√2))
Check all details, there can be mistakes.