### Differential equations and Laplace transforms: Laplace-transformations

### Laplace transforms of differential equations

The Laplace transform of the derivative #f'# of a function #f# can be expressed in the Laplace transform of #f# without the use of derivatives:

Derivative in the time domain If #f# is a differentiable function on #\ivco{0}{\infty}#, the Laplace transform #\laplace{f}# exists, and #\lim_{t\to\infty}\ee^{-st} f(t) = 0#, then \[\mathcal{L} f' (s) = s\cdot (\mathcal{ L}f)(s)-f(0)\]

More generally, if #n# is a natural number, #f# is a piecewise #n#-fold differentiable function, and, for all #k# with #0\le k\lt n#, the limit#\lim_{t\to\infty}\ee^{-st} f^{(k)}(t) = 0# exists, then \[ \laplace{\left(f^{(n)}\right)}(s) = s^n\cdot \laplace{f}(s)-s^{n-1}f(0)-s^{n-2}f'(0)-\cdots-s\cdot f^{(n-2)}(0)-f^{(n-1)}(0)\]

Thanks to this property we can convert a linear differential equation with constant coefficients using Laplace transform into an algebraic equation. By calculating the inverse Laplace transform we can solve the differential equation. For a second order ODE are the operations shown below in a diagram

\[\begin{array}{lcr}f''(t)+p\cdot f'(t)+q = r(t)&{\laplace{}\atop\longrightarrow}& P(s)\cdot\laplace{f}(s) = \laplace{r}(s)\\ \text{solution of }\big\uparrow&&\big\downarrow\text{ solve}\\ f(t) =\mathcal{L}^{-1}(F)(t)&{\mathcal{L}^{-1}\atop \longleftarrow}& \laplace{f}(s) = F(s)\end{array}\]

Below are examples of this solution method.

#x(t) =# # {{\sin \left(t\right)}\over{2}}-{{t\cdot \cos \left(t\right)}\over{2}} #

In order to find the solution, we write #y=\mathcal{L}(x)#. Then

\[\begin{array}{rcl}

\mathcal{L}(x'') (s)&=& s^2\cdot y(s)-2s =s^2\cdot y(s) \\

\mathcal{L}(-\sin \left(t\right)) (s)&=&\displaystyle -{{1}\over{s^2+1}} \\

\end{array}\]

Therefore, after all terms are moved to the left, the Laplace transform applied to the differential equation \( {{d^2}\over{d t^2}} x+x=\sin(t)\) gives

\[s^2\cdot y+y-{{1}\over{s^2+1}}=0\]

This can be rewritten to

\[\left(s^2+1\right)\cdot y-{{1}\over{s^2+1}}=0\]

Solving this equation with unknown #y# gives

\[ y (s)= {{1}\over{s^4+2\cdot s^2+1}} \]

Partial fraction decomposition of the right-hand side leads to

\[ \begin{array}{rcl}y(s)

&=&\displaystyle {{1}\over{\left(s^2+1\right)^2}}\\

\end{array}\]

so determining the inverse Laplace transform gives:

\[ x (t)= {{\sin \left(t\right)}\over{2}}-{{t\cdot \cos \left(t\right)}\over{2}} \]

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