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Use the Laplace transform?

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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Maurizio T. | Statistics Ph.D and CFA charterholder with a true passion to teach.Statistics Ph.D and CFA charterholder wi...
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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'') = s2Y(s) - c0s-c1, L(y') = sY(s) - c0 , L(y) = Y(s), and L(0)=0  where c0 = y(0)=1, c1=y'(0)= -1 and where Y(s) is the Laplace transform.

Hence you get the equation:

(s2Y(s)-1s+1)-(sY(s)-1) -6Y(s) = 0 

which you can semplify to

s2Y(s)-sY(s)-6Y(s) = s-2


Y(s) = (s-2)/(s2-s-6) = s/(s2-s-6) - 2/(s2-s-6)


the solution is given by y= L-1(Y(s)) = L-1(s/(s2-s-6)) - L-1(2/(s2-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) = ∫0y(t) e-stdt              (1)

 Laplace transforms of derivatives can be found using integration by parts. We have

 F1(s) = L y'(t) = ∫0 y'(t) e-stdt  = ∫0 e-st d y(t) =

          y(t) e-st (from 0 to ∞) +s ∫0 y(t) e-st dt


                                          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

   F2(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)/(s2 -s -6)   (4)

The denominator can be factored: 

                         s2 - 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)


             (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)( e3t + 4e-2t)