Hard Inverse Laplace Transform???

Question

i had this problem in my Differential Equation Exam today, and i was unable to solve cause i apply many times the convolution theorem and the answer given it wasn't the same

$$y''-5y=f(t)\qquad f(t) = \begin{cases} t^2 & 0\le t<1 & \\ 0&t\ge 1 \end{cases}\qquad \left\{ \begin{array}{c} y(0)=0\\ y'(0)=5 \\ \end{array} \right. $$ from here I found $Y(s)$ and when i try to solve it, my answer involves higher radicals numbers and exponentials because the term $s^2-5$ which suggest $\sinh(\sqrt{5}t)$. $$Y(s)=\frac{5s^3-2-e^{-s}(s^2+2s+2)}{s^3(s^2-5)}$$ The answer is: $$y(t)=-\frac{1}{125}(2+10t+25t^2-2e^{5t})u(t)+\frac{1}{250}(45+5t+25t^2-72e^{5(t-1)})u(t-1)$$

Answer 1

I don't think your given solution matches up with the equation. Either the exponential should be $e^{\sqrt{5}t}$ instead of $e^{\sqrt5 t}$, or the equation should be $y'' - 25y$ instead. Ask your instructor.

I'll provide my answer to the original equation. Rewrite $f(t)$ as

$$ f(t) = t^2\big[u(t)-u(t-1)\big] = t^2u(t) - \big[(t-1)^2 + 2(t-1) + 1\big]u(t-1) $$

Then

$$ F(t) = \frac{2}{s^3} - \left[\frac{2}{s^3} + \frac{2}{s^2} + \frac{1}{s}\right]e^{-s} = \frac{2 - (s^2+2s+2)e^{-s}}{s^3} $$

Transforming the entire equation

$$ (s^2-5)Y - 5 = \frac{2 - (s^2+2s+2)e^{-s}}{s^3} $$

The solution in the Laplace domain is

$$ Y(s) = \frac{5}{s^2-5} + \frac{2}{s^3(s^2-5)} - \frac{s^2+2s+2}{s^3(s^2-5)} e^{-s} $$

You now have two fractions to decompose. Use any method you like.

Here's an "easy" way without involving systems of equations (you don't have to do this)

\begin{align} \frac{25}{s^3(s^2-5)} &= \frac{s^4 - s^2(s^2-5) - 5(s^2-5)}{s^3(s^2-5)} &&= \frac{s}{s^2-5} - \frac{1}{s} - \frac{5}{s^3} \tag{1} \\ \frac{5s}{s^3(s^2-5)} &= \frac{s^3 - s(s^2-5)}{s^3(s^2-5)} &&= \frac{1}{s^2-5} - \frac{1}{s^2} \tag{2} \\ \frac{5s^2}{s^3(s^2-5)} &= \frac{s^4 - s^2(s^2-5)}{s^3(s^2-5)} &&= \frac{s}{s^2-5} - \frac{1}{s} \tag{3} \end{align}

Adding $(1)$, $(2)$, $(3)$

\begin{align} \frac{2}{s^3(s^2-5)} &= \frac{2}{25}\left[\frac{s}{s^2-5} - \frac{1}{s} - \frac{5}{s^3}\right] \\ \frac{s^2+2s+2}{s^3(s^2-5)} &= \frac15\left[\frac{s}{s^2-5} - \frac{1}{s} \right] + \frac{2}{5}\left[\frac{1}{s^2-5} - \frac{1}{s^2}\right] + \frac{2}{25}\left[\frac{s}{s^2-5} - \frac{1}{s} - \frac{5}{s^3}\right] \\ &= \frac{7}{25}\frac{s}{s^2-5} + \frac{2}{5}\frac{1}{s^2-5} - \frac{7}{25}\frac{1}{s} - \frac{2}{5}\frac{1}{s^2} - \frac{2}{5}\frac{1}{s^3} \end{align}

You can further decompose the $(s^2-5)$ fractions, but I'll use $\sinh$ and $\cosh$ here. Reverse transforming gives

\begin{align} y(t) &= \left[\sqrt{5} \sinh\big(\sqrt{5}t\big) + \frac{2}{25}\cosh\big(\sqrt{5}t\big) - \frac{2}{25} - \frac{1}{5}t^2\right]u(t) \\ &\ - \left[\frac{2}{5\sqrt{5}}\sinh\big(\sqrt{5}(t-1)\big) + \frac{7}{25}\cosh\big(\sqrt{5}(t-1)\big) \\ - \frac{7}{25} - \frac{2}{5}(t-1)-\frac{1}{5}(t-1)^2 \right]u(t-1) \end{align}

Answer 2

$$ f(t) = t^2(u(t)-u(t-1)) $$

then after transforming

$$ s^2 Y(s)-5Y(s) = s \dot y_0 + y_0 + \frac{2}{s^3}-\frac{s^2+2s+2}{s^3}e^{-s} $$

and then

$$ Y(s) = \frac{1}{s^2-5}\left(s \dot y_0 + y_0 \right)+\frac{2}{s^3} \frac{1}{s^2-5}-\frac{1}{s^3} \frac{1}{s^2-5}\left(s^2+2s+2\right)e^{-s} $$

now using the transform tables

$$ \frac{a}{s^2-a^2}\leftrightarrow \sinh(at)\\ \frac{n!}{(s-a)^{n+1}}\leftrightarrow t^ne^{at}\\ e^{-c s}F(s)\leftrightarrow f(t-c)u(t-c) $$

etc.

NOTE

If $\frac{1}{s^3} \frac{1}{s^2-5}\leftrightarrow g(t)$ then $ s^n\frac{1}{s^3} \frac{1}{s^2-5}\leftrightarrow \frac{d^n}{dt^n}g(t)$