Use the Laplace transform to solve y"-y'-6y=0; y(0)=1, y'(0)=-1.
Answer: y=(1/5)(e^(3t)+4e^(-2t))
I don't know how to take the Laplace transform for both sides. Help me step by step.
Use the Laplace transform to solve y"-y'-6y=0; y(0)=1, y'(0)=-1.
Answer: y=(1/5)(e^(3t)+4e^(-2t))
I don't know how to take the Laplace transform for both sides. Help me step by step.
Sun,
Laplace transforms are used to solve initial-value problems given b n-th order linear differential equations with constant coefficients. In your case n=2. If L indicates the Laplace transform operator, we can write the assigned equation as
L(y'') - L(y') -6L(y) = L(0).
I am sure your textbook shows that L(y'') = s^{2}Y(s) - c_{0}s-c_{1}, L(y') = sY(s) - c_{0} , L(y) = Y(s), and L(0)=0 where c_{0} = y(0)=1, c_{1}=y'(0)= -1 and where Y(s) is the Laplace transform.
Hence you get the equation:
(s^{2}Y(s)-1s+1)-(sY(s)-1) -6Y(s) = 0
which you can semplify to
s^{2}Y(s)-sY(s)-6Y(s) = s-2
or
Y(s) = (s-2)/(s^{2}-s-6) = s/(s^{2}-s-6) - 2/(s^{2}-s-6)
Hence
the solution is given by y= L^{-1}(Y(s)) = L^{-1}(s/(s^{2}-s-6)) - L^{-1}(2/(s^{2}-s-6)).
Therefore, to complete the problem you need to compute the two inverse Laplace transforms. I suppose that you were taught that, right?
Let F(s) be the Laplace transform of y(t). Thus
Ly(t) = ∫_{0}^{∞ }y(t) e-^{st}dt (1)
Laplace transforms of derivatives can be found using integration by parts. We have
F_{1}(s) = L y'(t) = ∫_{0}^{∞} y'(t) e^{-st}dt = ∫_{0}^{∞} e^{-st} d y(t) =
y(t) e^{-st} (from 0 to ∞) +s ∫_{0}^{∞} y(t) e^{-st} dt
or
Ly'(t) = sF(s) -1 (2)
after using initial condition for y(t) and the fact that e^{-st} = 0 for t ↔∞.
By following the same steps we can show that fhe Laplace transform for the secpnd derivative
F_{2}(s) = Ly''(t) = sF1(s) + 1 = s[sF(s) - 1] -(-1) (3)
The last term in (3) is just y'(0) = -1. By taking Laplace transforms of all terms in the left side of
your equation and making use of expressions from (1) to (3), we obtain:
F(s) (s2 - s -6) - s + 2 = 0 or F(s) = (s -2)/(s^{2} -s -6) (4)
The denominator can be factored:
s^{2} - s - 6 = (s+2)(s-3)
Thus we have
F(s) = (s-2)/[s+2)(s-3)]= (1/5) (s-2)[ 1/(s-3) - 1/(s+2)] (5)
If you notice that
(s-2)/(s-3) = (s-2-1+1)/(s-3) = 1 + 1/(s-3)
and
(s-2)/(s+2) = (s-2 +4 -4)/(s+2) = 1 - 4/(s+2)
then you will come up with the following expression for F(s):
F(s) = (1/5) [1/(s-3) + 4/(s+2)] (6)
From previous tutoring sections we already know that the Laplace transform of e ^{at} is 1/(s-a)
where a = 3 and a = -2 in in this problem. Thus, taking inverse transforms we will come up with
the final answer:
y(t) = (1/5)( e^{3t} + 4e^{-2t})