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

Apr 5, 2018

See below.

#### Explanation:

Considering

$y ' + a y = b \sin \left(\omega t\right)$ we have

$L \left[y ' + a y\right] = L \left[b \sin \left(\omega t\right)\right]$ or

$s Y \left(s\right) - y \left(0\right) + a Y \left(s\right) = b \frac{\omega}{{\omega}^{2} + {s}^{2}}$ or

$Y \left(s\right) \left(s + a\right) = b \frac{\omega}{{\omega}^{2} + {s}^{2}} + y \left(0\right)$ or

$Y \left(s\right) = \frac{b \omega}{\left(s + a\right) \left({s}^{2} + {\omega}^{2}\right)} + \frac{y \left(0\right)}{s + a}$

and after inversion

$y \left(t\right) = \left(\frac{b}{{\omega}^{2} + {a}^{2}} \left(a \sin \left(\omega t\right) - \omega \cos \left(\omega t\right)\right) + \left(y \left(0\right) + \frac{b \omega}{{\omega}^{2} + {a}^{2}}\right) {e}^{- a t}\right) u \left(t\right)$

where $u \left(t\right)$ is the unit step function

Apr 5, 2018

$y \left(t\right) = 1.25 {e}^{- 5.2 t} + 3.25 \sin \left(2 t\right) - 1.25 \cos \left(2 t\right)$

#### Explanation:

We seek a solution of the IVP Differential Equation:

$y ' + 5.2 y = 19.4 \sin \left(2 t\right)$ with $y \left(0\right) = 0$

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:

$s F \left(s\right) - f \left(0\right) + 5.2 F \left(s\right) = \left(19.4\right) \frac{2}{{s}^{2} + {2}^{2}}$

And we use the IV's and re-arrange for $F \left(s\right)$:

$s F \left(s\right) - 0 + 5.2 F \left(s\right) = \frac{38.8}{{s}^{2} + 4}$

$\therefore \left(s + 5.2\right) F \left(s\right) = \frac{38.8}{{s}^{2} + 4}$

$\therefore F \left(s\right) = \frac{38.8}{\left({s}^{2} + 4\right) \left(s + 5.2\right)}$

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:

$\frac{38.8}{\left({s}^{2} + 4\right) \left(s - 5.2\right)} \equiv \frac{A s + B}{{s}^{2} + 4} + \frac{C}{s + 5.2}$

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:

$f \left(t\right) = 6.5 \setminus \frac{1}{2} \sin \left(2 t\right) - 1.25 \setminus \cos \left(2 t\right) + 1.25 \setminus {e}^{- 5.2 t}$

Hence, the solution ot the IVP is:

$y \left(t\right) = 3.25 \setminus \sin \left(2 t\right) - 1.25 \cos \left(2 t\right) + 1.25 {e}^{- 5.2 t}$