Question

Laplace Transforms y'+5.2y=19.4sin(2t) By:y(0)=0 y=?

Answer 1

See below.

Explanation:

Considering

`y'+ay= b sin(omega t)` we have

`L[y'+ay] = L[ b sin(omega t)]` or

`sY(s)-y(0)+aY(s) = b omega/(omega^2+s^2)` or

`Y(s)(s+a) = b omega/(omega^2+s^2)+y(0)` or

`Y(s) = (b omega)/((s+a)(s^2+omega^2))+(y(0))/(s+a)`

and after inversion

`y(t) = (b/(omega^2+a^2)(a sin(omega t)-omega cos(omega t))+(y(0)+(b omega)/(omega^2+a^2))e^(-a t) )u(t)`

where `u(t)` is the unit step function

Answer 2

`y(t) = 1.25 e^(-5.2t) + 3.25 sin(2t) -1.25 cos(2t)`

Explanation:

We seek a solution of the IVP Differential Equation:

`y' + 5.2y=19.4sin(2t)` with `y(0)=0`

Using laplace Transformations.

We will need the following standard Laplace transforms 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),), (e^(at), 1/(s-a), a " constant"), (sinat, a/(s^2+a^2), a " constant"), (cosat, s/(s^2+a^2), a " constant") :} #

Then, taking Laplace transforms of the given equation, exploiting linearity we have:

`ℒ{y'} + 5.2ℒ{y} = 19.4ℒ{sin(2t)}`

Using the standard results, we have:

`sF(s) -f(0) + 5.2F(s) = (19.4 )2/(s^2+2^2)`

And we use the IV's and re-arrange for `F(s)`:

`sF(s) -0 + 5.2F(s) = (38.8)/(s^2+4)`

`:. (s+5.2)F(s) = (38.8)/(s^2+4)`

`:. F(s) = (38.8)/((s^2+4)(s+5.2))`

As if often the case with a LT solution, we gain an advantage by transforming a DE into an algebraic equation at the expense of taking inverses transformations. In preparation for this, we will need to decompose the composite fraction into partial fractions:

`(38.8)/((s^2+4)(s-5.2)) -= (As+B)/(s^2+4) + C/(s+5.2)`

Using the cover-up method and comparing coefficients, we ascertain that:

`A = -1.25`, `B=6.5` and `C=1.25`

Thus we can gain the solution of the DE by taking inverse laplace transformations:

`f(t) = ℒ^(-1){F(s)}`

`\ \ \ \ \ \ = ℒ^(-1){(6.5-1.25s)/(s^2+4) + 1.25/(s+5.2)}`

`\ \ \ \ \ \ = 6.5 \ ℒ^(-1){1/(s^2+4)} -1.25 \ ℒ^(-1){s/(s^2+4)} + 1.25 \ ℒ^(-1){1/(s+5.2)}`

And again using the table of transformation we have:

`f(t) = 6.5 \ 1/2sin(2t) -1.25 \ cos(2t) + 1.25 \ e^(-5.2t)`

Hence, the solution ot the IVP is:

`y(t) = 3.25 \ sin(2t) -1.25 cos(2t) + 1.25 e^(-5.2t)`