# Jun's question via email about Laplace Transform

• MHB
• Prove It
In summary, the solution to the initial value problem using Laplace Transforms is given by $y(t) = 2\cos(2t) + c_1\sin(2t) + \cos(2(t-6))H(t-6)$ where $c_1 = \frac{c}{4}$.
Prove It
Gold Member
MHB
$\displaystyle y\left( t \right)$ satisfies the initial value problem

$\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}t^2} + 4\,y= -8\,H\left( t - 6 \right) , \quad y\left( 0 \right) = 2 , \,\, y'\left( 0 \right) = 0$

Find the solution to the initial value problem using Laplace Transforms.

Upon taking the Laplace Transform of the equation we have

\displaystyle \begin{align*} s^2\,Y\left( s \right) - s\,y\left( 0 \right) - y'\left( 0 \right) + 4\,Y\left( s \right) &= -\frac{8\,\mathrm{e}^{-6\,s}}{s} \\ s^2 \,Y\left( s \right) - 2\,s - 0 + 4\,Y\left( s \right) &= -\frac{8\,\mathrm{e}^{-6\,s}}{s}\\ \left( s^2 + 4 \right) Y\left( s \right) - 2\,s &= -\frac{8\,\mathrm{e}^{-6\,s}}{s} \\ \left( s^2 + 4\right) Y\left( s \right) &= 2\,s - \frac{8\,\mathrm{e}^{-6\,s}}{s} \\ Y\left( s \right) &= \frac{2\,s}{s^2 + 4} - \frac{8\,\mathrm{e}^{-6\,s}}{s\left( s^2 + 4 \right) } \\ Y\left( s \right) &= 2 \left[ \frac{s}{s^2 + 4} - \frac{4\,\mathrm{e}^{-6\,s}}{s\left( s^2 + 4 \right) } \right] \end{align*}

The first term's Inverse Transform can be read off the tables. The second requires the second shift theorem: $\displaystyle \mathcal{L}\,\left\{ f\left( t - a \right) \, H\left( t - a \right) \right\} = \mathrm{e}^{-a\,s}\,F\left( s \right)$.

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

Applying Partial Fractions:

\displaystyle \begin{align*} \frac{A}{s} + \frac{B\,s + C}{s^2 + 4} &\equiv \frac{4}{s\left( s^2 + 4 \right) } \\ A\left( s^2 + 4 \right) + \left( B\,s + C \right) s &\equiv 4 \end{align*}

Let $\displaystyle s = 0 \implies 4\,A = 4 \implies A = 1$, then

\displaystyle \begin{align*} 1\left( s^2 + 4 \right) + \left( B\,s + C \right) s &\equiv 4 \\ s^2 + 4 + B\,s^2 + C\,s &\equiv 4 \\ \left( B + 1 \right) s^2 + C\,s + 4 &\equiv 0\,s^2 + 0\,s + 4 \end{align*}

It's clear that $\displaystyle B + 1 = 0 \implies B = -1$ and $\displaystyle C = 0$. Thus

\displaystyle \begin{align*} F\left( s \right) &= \frac{1}{s} - \frac{s}{s^2 + 4} \\ f\left( t \right) &= 1 - \cos{ \left( 2\,t \right) } \\ f\left( t - 6 \right) \, H\left( t - 6 \right) &= \left\{ 1 - \cos{ \left[ 2 \left( t - 6 \right) \right] } \right\} \, H\left( t - 6 \right) \end{align*}

So from our original DE

\displaystyle \begin{align*} Y\left( s \right) &= 2 \left[ \frac{s}{s^2 + 4} - \frac{4\,\mathrm{e}^{-6\,s}}{s\left( s^2 + 4 \right) } \right] \\ \\ y \left( t \right) &= 2\left[ \cos{ \left( 2\,t \right) } - \left\{ 1 - \cos{ \left[ 2\left( t - 6 \right) \right] } \right\} \, H\left( t - 6 \right) \right] \\ &= 2 \left[ \cos{ \left( 2\,t \right) } + \left\{ \cos{ \left[ 2\left( t - 6 \right) \right] } - 1 \right\} \, H\left( t - 6 \right) \right] \end{align*}

benorin
Was it given that ##y’(0)=0##? The second line after you apply the transform (the 0), I think this should be replaced by an arbitrary constant, say c? Try that. Since it’s a second order IVP with only one initial value given the solution should contain an unknown constant.

Here’s you work accounting for this constant up to a certain point with differences ##\boxed{\text{new term(s)}}##.

Your work:
Upon taking the Laplace Transform of the equation we have

\begin{align*} s^2\,Y\left( s \right) - s\,y\left( 0 \right) - y'\left( 0 \right) + 4\,Y\left( s \right) &=& -\frac{8\,\mathrm{e}^{-6\,s}}{s} \\ s^2 \,Y\left( s \right) - 2\,s - \boxed{c} + 4\,Y\left( s \right) &=& -\frac{8\,\mathrm{e}^{-6\,s}}{s}+\boxed{c} \\ \left( s^2 + 4 \right) Y\left( s \right) - 2\,s &=& -\frac{8\,\mathrm{e}^{-6\,s}}{s}+\boxed{c} \\ \left( s^2 + 4\right) Y\left( s \right) &=& 2\,s - \frac{8\,\mathrm{e}^{-6\,s}}{s}+\boxed{c} \\ Y\left( s \right) &=& \frac{2\,s}{s^2 + 4} - \frac{8\,\mathrm{e}^{-6\,s}}{s\left( s^2 + 4 \right) } +\boxed{\frac{c}{s^2+4}} \\ Y\left( s \right) &=& 2 \left[ \frac{s}{s^2 + 4} - \frac{4\,\mathrm{e}^{-6\,s}}{s\left( s^2 + 4 \right) } + \boxed{\frac{c}{2(s^2+4)}} \right] \\ \end{align*}

Last edited:
Sorry had latex problems, refresh the page. Simple from there, just take the inverse transform of the last boxed term. Should be ##c_1 \sin (2t)## where ##c_1 = \tfrac{c}{4}##. Got it from there?

## 1. What is a Laplace Transform?

A Laplace Transform is a mathematical operation that converts a function of time into a function of complex frequency. It is often used in engineering and physics to solve differential equations and analyze systems in the frequency domain.

## 2. How is a Laplace Transform calculated?

The Laplace Transform is calculated by integrating the function of time multiplied by the exponential of negative time. This integral is evaluated from 0 to infinity. The result is a function of complex frequency.

## 3. What are the applications of Laplace Transform?

Laplace Transform has many applications in engineering and physics, such as solving differential equations, analyzing control systems, and studying circuits and signals in the frequency domain. It is also used in image processing, economics, and other fields.

## 4. What are the advantages of using Laplace Transform?

Laplace Transform allows for the analysis of complex systems in the frequency domain, which can provide insights that are not easily obtained in the time domain. It also simplifies the solution of differential equations and allows for the use of algebraic methods instead of calculus.

## 5. Are there any limitations to using Laplace Transform?

One limitation of Laplace Transform is that it only works for functions that are defined for all positive time. It also cannot be used for functions that have an infinite number of discontinuities or are not piecewise continuous. Additionally, the inverse Laplace Transform may not have a closed-form solution, making it difficult to find the original function.

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