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

2 Answers
Apr 5, 2018

Answer:

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

Apr 5, 2018

Answer:

# 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) #