# Can u guys helpe please? Ive been trying to solve two questions and i simply have no luck. They are: Inverse laplace of s/(s^4+4a^4) And Inverse laplace of s/(s^4+s^2+1) If you guys could gelp me I wpuld be grateful

Mar 13, 2017

 ℒ^-1 {s/(s^4+4a^4)} \ \ \ \ \ = (sin(at)sinh(at))/(2a^2)
 ℒ^-1 {s/(s^4+s^2+1)} = 2/sqrt(3)sin(sqrt(3)/2)sinh(1/2t)

#### Explanation:

Part (A)

First we can complete the square:

$\frac{s}{{s}^{4} + 4 {a}^{4}} = \frac{s}{{\left({s}^{2} + 2 {a}^{2}\right)}^{2} - 4 {a}^{2} {s}^{2}}$
$\text{ } = \frac{s}{{\left({s}^{2} + 2 {a}^{2}\right)}^{2} - {\left(2 a s\right)}^{2}}$

We now have the difference of two squares, so:

$\frac{s}{{s}^{4} + 4 {a}^{4}} = \frac{s}{\left({s}^{2} + 2 {a}^{2} + 2 a s\right) \left({s}^{2} + 2 {a}^{2} - 2 a s\right)}$

Now let us form the partial fraction decomposition for the expression, which will be of the form:

$\frac{s}{{s}^{4} + 4 {a}^{4}} = \frac{A}{{s}^{2} + 2 {a}^{2} + 2 a s} + \frac{B}{{s}^{2} + 2 {a}^{2} - 2 a s}$
$\text{ } = \frac{A \left({s}^{2} + 2 {a}^{2} - 2 a s\right) + B \left({s}^{2} + 2 {a}^{2} + 2 a s\right)}{\left({s}^{2} + 2 {a}^{2} + 2 a s\right) \left({s}^{2} + 2 {a}^{2} - 2 a s\right)}$

$\therefore s = A \left({s}^{2} + 2 {a}^{2} - 2 a s\right) + B \left({s}^{2} + 2 {a}^{2} + 2 a s\right)$

Equating Coefficients we get:

$C o e f f \left(s\right) \setminus \setminus \implies 1 = - 2 a A + 2 a B \setminus \setminus \setminus \setminus \setminus \ldots . . \left[1\right]$
$C o e f f \left({s}^{2}\right) \implies 0 = A + B \setminus \setminus \setminus \setminus \setminus \ldots . . \left[2\right]$

Now we solve these simultaneous equations:

$E q \left[1\right] + \left(2 a\right) E q \left[2\right] \implies 1 = 4 a B$
$\therefore A = - \frac{1}{4 a} , B = \frac{1}{4 a}$

Se we can write the original expression as:

$\frac{s}{{s}^{4} + 4 {a}^{4}} = \frac{1}{4 a} \left\{\frac{1}{{s}^{2} + 2 {a}^{2} - 2 a s} - \frac{1}{{s}^{2} + 2 {a}^{2} + 2 a s}\right\}$
$\text{ } = \frac{1}{4 a} \left\{\frac{1}{{\left(s - a\right)}^{2} - {a}^{2} + 2 {a}^{2}} - \frac{1}{{\left(s + a\right)}^{2} - {a}^{2} + 2 {a}^{2}}\right\}$
$\text{ } = \frac{1}{4 a} \left\{\frac{1}{{\left(s - a\right)}^{2} + {a}^{2}} - \frac{1}{{\left(s + a\right)}^{2} + {a}^{2}}\right\}$

And we can now take inverse Laplace Transforms using standard results to get:

 ℒ^-1 {s/(s^4+4a^4)} = 1/(4a){e^(at)sin(at)/a - e^(-at)sin(at)/a}
$\text{ } = \frac{1}{4 {a}^{2}} \sin \left(a t\right) \left\{{e}^{a t} - {e}^{- a t}\right\}$

$\text{ } = \frac{1}{4 {a}^{2}} \sin \left(a t\right) 2 \sinh \left(a t\right)$

$\text{ } = \frac{\sin \left(a t\right) \sinh \left(a t\right)}{2 {a}^{2}}$

Part (B)
In a similar way to the first solution we want to make the denominator the difference of two squares so that we have two factors leading to two partial fractions. If we complete the square on this denominator the method fails as we get

${s}^{4} + {s}^{2} + 1 = {\left({s}^{2} + \frac{1}{2}\right)}^{2} + \frac{3}{4}$

Which is not the difference of two squares; so let's try a little manipulation:

$\frac{s}{{s}^{4} + {s}^{2} + 1} = \frac{s}{\left({s}^{4} + 2 {s}^{2} + 1\right\} - {s}^{2}}$
$\text{ } = \frac{s}{{\left({s}^{2} + 1\right)}^{2} - {s}^{2}}$
$\text{ } = \frac{s}{\left({s}^{2} + 1 - s\right) \left({s}^{2} + 1 + s\right)}$

Again we find the partial fraction decomposition; which will be of the form;:

$\frac{s}{{s}^{4} + {s}^{2} + 1} = \frac{A}{{s}^{2} + 1 - s} + \frac{B}{{s}^{2} + 1 + s}$
$\text{ } = \frac{A \left({s}^{2} + 1 + s\right) + B \left({s}^{2} + 1 - s\right)}{\left({s}^{2} + 1 - s\right) \left({s}^{2} + 1 + s\right)}$

$\therefore s = A \left({s}^{2} + 1 + s\right) + B \left({s}^{2} + 1 - s\right)$

Equating Coefficients we get:

$C o e f f \left(s\right) \setminus \setminus \implies 1 = A - B \setminus \setminus \setminus \setminus \setminus \ldots . . \left[1\right]$
$C o e f f \left({s}^{2}\right) \implies 0 = A + B \setminus \setminus \setminus \setminus \setminus \ldots . . \left[2\right]$

Now we solve these simultaneous equations:

$E q \left[1\right] + E q \left[2\right] \implies 1 = 2 A$
$\therefore A = \frac{1}{2} , B = - \frac{1}{2}$

Se we can write the original expression as:

$\frac{s}{{s}^{4} + {s}^{2} + 1} = \frac{1}{2} \left\{\frac{1}{{s}^{2} + 1 - s} - \frac{1}{{s}^{2} + 1 + s}\right\}$
$\text{ } = \frac{1}{2} \left\{\frac{1}{{\left(s - \frac{1}{2}\right)}^{2} - {\left(\frac{1}{2}\right)}^{2} + 1} - \frac{1}{{\left(s + \frac{1}{2}\right)}^{2} - {\left(\frac{1}{2}\right)}^{2} + 1}\right\}$
$\text{ } = \frac{1}{2} \left\{\frac{1}{{\left(s - \frac{1}{2}\right)}^{2} - \frac{3}{4}} - \frac{1}{{\left(s + \frac{1}{2}\right)}^{2} - \frac{3}{4}}\right\}$
$\text{ } = \frac{1}{2} \left\{\frac{1}{{\left(s - \frac{1}{2}\right)}^{2} - {\left(\frac{\sqrt{3}}{2}\right)}^{2}} - \frac{1}{{\left(s + \frac{1}{2}\right)}^{2} - {\left(\frac{\sqrt{3}}{2}\right)}^{2}}\right\}$

And we can now take inverse Laplace Transforms using standard results to get:

 ℒ^-1 {s/(s^4+s^2+1)} = 1/2{e^(1/2t)sin(sqrt(3)/2)/(sqrt(3)/2)) - e^(-1/2t)sin(sqrt(3)/2)/(sqrt(3)/2))}
$\text{ } = \frac{1}{2} \cdot \frac{2}{\sqrt{3}} \sin \left(\frac{\sqrt{3}}{2}\right) \left\{{e}^{\frac{1}{2} t} - {e}^{- \frac{1}{2} t}\right\}$
$\text{ } = \frac{1}{\sqrt{3}} \sin \left(\frac{\sqrt{3}}{2}\right) 2 \sinh \left(\frac{1}{2} t\right)$
$\text{ } = \frac{2}{\sqrt{3}} \sin \left(\frac{\sqrt{3}}{2}\right) \sinh \left(\frac{1}{2} t\right)$