# Differential equations using Laplace transforms

By using Laplace transforms find $x(t)$ from the coupled differential equations$$\frac{dx}{dt} = -kx+gy+E,$$$$\frac{dy}{dt} = -ky-gx,$$for some functions $x(t),$ and $y(t)$, where $E, k, g$ are real. And we have the condition $x(t) = y(t) = 0$ for $t \leq 0.$

Attempt:

I know that the solution must be:

$$x(t)=\frac{Ek}{k^2+g^2} \Big[ 1+ e^{-kt} \Big[ -\cos(gt)+ \frac{g}{k} \sin(gt) \Big] \Big]. \tag{1}$$

I am trying to arrive at equation (1). Do I need to be considering the steady-state of the system? If so, when I set the two equations to zero, how can I take the Laplace transform of both sides? Because we do not have an explicit expression given for the functions $x(t)$ and $y(t).$

Also we don't know if the Laplace transforms of $x(t)$ and $y(t)$ are convergent in the region $Real (s) > 0.$

Any explanation would be appreciated.

• You don't need to consider steady-state of the system, simply take the Laplace transform of the coupled ODEs and solve for $X(s)$, where $X(s)$ is the Laplace transform of $x(t)$. – Chee Han May 18 '16 at 12:46
• Also note that you have defined the starting conditions and derivatives so there are no alternate solutions to the evolution from then on. e.e you are evolving the condition x,y=0,0 and dx/dt=E at t=0 so it's all smooth sailing; so to say. – rrogers May 18 '16 at 14:39

Denote the Laplace transform of $x(t)$ and $y(t)$ by $X(s)$ and $Y(s)$. Taking Laplace transform of the coupled ODEs yields the following: \begin{align} sX & = -kX + gY + \dfrac{E}{s} \tag{1}\\ sY & = -kY - gX \tag{2} \end{align} where we use the assumption that we have zero initial conditions. Rearranging (2) to get an expression for $Y(s)$: \begin{align*} (s+k)Y & = -gX\\ \implies Y & = -\dfrac{gX}{s+k}\\ \end{align*} Substituting this expression into (1) gives: \begin{align*} sX + kX & = g\left(-\dfrac{gX}{s+k}\right) + \dfrac{E}{s}\\ \left(s+k+\dfrac{g^2}{s+k}\right)X & = \dfrac{E}{s}\\ \left(\dfrac{(s+k)^2+g^2}{s+k}\right)X & = \dfrac{E}{s}\\ X(s) & = \dfrac{E(s+k)}{s[(s+k)^2+g^2]}\\ & = E\left(\dfrac{A}{s} + \dfrac{Bs+C}{(s+k)^2+g^2}\right)\\ & = EA\left\{\dfrac{1}{s}\right\} + EB\left\{\dfrac{s+k}{(s+k)^2+g^2}\right\} + \dfrac{E(C-Bk)}{g}\left\{\dfrac{g}{(s+k)^2+g^2}\right\} \end{align*} where \begin{equation} A = \dfrac{k}{k^2+g^2},\quad B = -\dfrac{k}{k^2+g^2},\quad C = \dfrac{g^2-k^2}{k^2+g^2} \end{equation} You should be able to figure out how to invert them; the frequency shifting theorem will be helpful.
• Thank you so much for your input. But how do we know that the Laplace transforms $X(s)$ and $Y(s)$ are absolutely convergent in the region $>0$? – Merin May 19 '16 at 6:40
• Not necessarily. $1/s$ is convergent for all $\Re(s)>0$, but the other two terms are convergent only for all $\Re(s)>-k$. So in order for $X(s)$ to converge in the region you wanted, $k$ should be a nonnegative real number. You can refer to wikipedia about the convergence region for different function here: en.wikipedia.org/wiki/… – Chee Han May 19 '16 at 8:51
• Well we know that $Y(s)=-gX/(s+k)$, so partial fraction decomposition will give us an extra term $D/s+k$, and this term converges for $\Re(s)>-k$. I don't think knowing the equilibrium solution of the system will tell us anything about the region of convergence in the Laplace space; I might be wrong but as far as I know, it seems like we can only tell the region of convergence when we have an explicit form of solution (either $x(t)$ or $X(s)$). Recall that we derive the region of convergence of $X(s)$ directly from the definition of Laplace transform. – Chee Han May 19 '16 at 12:16
• That was the right property. Since $\mathcal{L}^{-1}\{F(s)\} = \mathcal{L}^{-1}\left\{\dfrac{g}{s^2+g^2}\right\}=\sin(gt)$, we have that (from frequency shifting property) $\mathcal{L}^{-1}\{F(s+k)\} = \mathcal{L}^{-1}\left\{\dfrac{g}{(s+k)^2+g^2}\right\} = e^{-kt}\sin(gt)$. Note I didn't check the constants but you should be able to figure them out. Glad I can help (: – Chee Han May 19 '16 at 12:33