Question

Laplace transforms... How do I go about getting `y`? `y''-2y'-3y=0`, `y(4)=-3`, `y'(4)=-17`

Answer 1

I assume `y(0)=A` and `y'(0)=B`

After taking Laplace transform both sides,

`s^2*Y(s)-s*y(0)-y'(0)-2*[s*Y(s)-y(0)]-3Y(s)=0`

`s^2*Y(s)-As-B-2s*Y(s)-2A-3Y(s)=0`

`(s^2-2s-3)*Y(s)=As+2A+B`

`(s+1)*(s-3)*Y(s)=As+2A+B`

`Y(s)=(As+2A+B)/[(s+1)*(s-3)]`

`Y(s)=(5A+B)/4*1/(s+1)-(A+B)/4*1/(s-3)`

After taking inverse Laplace transformation both sides,

`y=(5A+B)/4*e^(-x)-(A+B)/4*e^(3x)`

Now, I use `y(4)=-3` condition,

`(5A+B)/4*e^(-4)-(A+B)/4*e^12=-3`

I differentiate both sides and apply `y'(4)=-17` condition,

`y'=-(5A+B)/4*e^(-x)-(3A+3B)/4*e^(3x)`

`-(5A+B)/4*e^(-4)-(3A+3B)/4*e^12=-17` or,

`(5A+B)/4*e^(-4)+(3A+3B)/4*e^12=17`

From these equations, `A=2e^4-5e^(-12)` and `B=25e^(-12)-2e^4`

Thus, `y=2e^4*e^(-x)-5e^(-12)*e^(3x)` or,

`y=2e^(4-x)-5e^(3x-12)`

Explanation:

1) I used Laplace transformation for `y(0)=A` and `y'(0)=B` conditions.

2) I solved `Y(s)`

3) I applied inverse Laplace transformation for finding `y`

4) I used `y(4)=-3` and `y'(4)=-17` conditions for finding A and B

Answer 2

`y(t) = 2e^(4-t) - 5e^(3t-12)`

Explanation:

We have:

`y''-2y'-3y=0` with `y(4)=-3, y'(4)=-17` ..... [A]

We note that this is a shifted value IVP so we cannot just take Laplace Transforms "as is", and we must perform a substitution to shift the initial conditions to that of an IVP.

We have initial conditions pivoting on `t=4`, so we need a `-4` IVP shift:

Let `tau = t-4 => t = tau+4`
And `u(tau)=y(tau+4) => y(t) = u(t-4)`

Then substituting into the DE [A] we get:

`u''-2u'-3u=0` with `u(0)=-3, u'(0)=-17` ..... [B]

We will need the following standard Laplace transform and inverses:

# {: (ul(f(t)=ℒ^(-1){F(s)}), ul(F(s)=ℒ{f(t)}), ul("Notes")), (f(t), F(s),), (f'(t), sF(s)-f(0),), (f''(t), s^2F(s)-s f(0)-s f'(0),), (e^(at), 1/(s-a),) :} #

We can now use this table to apply Laplace Transforms to [B]:

`ℒ \ {u''} - ℒ \ {2u'} -ℒ \ { 3u} = 0`

`:. s^2U-su'(0)-u(0) - 2{sU-s(0)} -3U = 0` where `U(s)= ℒ \ {u(t)}`

`:. (s^2U+3s+17) - 2(sU+3) -3U = 0`

`:. s^2U+3s+17 - 2sU-6 -3U = 0`

`:. (s^2-2s-3)U + 3s+11 = 0`

`:. (s-3)(s+1)U = -11-3s`
`:. U = (-11-3s)/((s-3)(s+1))`

Now we decompose into partial fractions:

`(-11-3s)/((s-3)(s+1)) -= a/(s-3)+b/(s+1)`
`" " -= (a(s+1) +b(s-3))/((s-3)(s+1))`
`:. -11-3s = a(s+1) +b(s-3)`

Put:

`s=3 => -20 = 4a => a=-5`
`s=-1 => -8 = -4b => b=2`

Thus:

`U(s) = -5/(s-3) +2/(s+1)`

And again using the above table we can now take inverse Laplace Transforms:

`u(tau) = -5e^(3tau) +2 e^(-tau)`
`\ \ \ \ \ \ = 2e^(-tau) - 5e^(3tau)`

And so, the solution we seek, after restoring the substitution is:

`y(t) = u(t-4)`
`\ \ \ \ \ \ = 2e^(-(t-4)) - 5e^(3(t-4))`
`\ \ \ \ \ \ = 2e^(4-t) - 5e^(3t-12)`

Side Note

Although this demonstrates how to deal with a shifted data IVPs, it is in fact much easier to solve this DE using traditional methods rather than Laplace Transforms, viz:

The associated Auxiliary equation is:

`m^2-2m-3 = 0 => (m-3)(m+1) = 0 => m=-1,3`

Yielding a general solution of the homogeneous equation of the form:

`y = Ae^(-t) + Be^(3t) => y' = -Ae^(-t) + 3Be^(3t)`

And using the initial condition `y(4) = -3, y'(4)=-17` we get:

`-3 = Ae^(-4) + Be^(12)` ..... [C]
`-17 = -Ae^(-4) + 3Be^(12)` ..... [D]

Solving Simultaneously, we get:

Eg [C]+ Eq [D]:

`-20 = 4Be^(12) => B = -5e^(-12)`

Subs `B` into [C]:

`-3 = Ae^(-4) -5e^12e^(-12)`
`:. Ae^(-4) = 5 - 3`
`:. Ae^(-4) = 2`
`:. A = 2e^(4)`

So the Particular Solution is:

`y = 2e^4e^(-t) -5e^(-12t)e^(3t)`
`\ \ \ = 2e^(4-t) -5e^(3t-12)`, as above