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.

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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?

Grigori S. | Certified Physics and Math Teacher G.S.Certified Physics and Math Teacher G.S.

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})

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