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I have used Laplace transforms to transform a system of 2 coupled second order ODEs into 2 simultaneous equations.

1st ode:

$$\frac{3d^2y}{dt^2}+\frac{dy}{dx}=0$$

2nd ode:

$$\frac{5d^2y}{dx^2}-\frac{dy}{dt}=0$$

y is a function of two variables $y(x,t)$

I define $L_t$ to be the Laplace transform wrt to $t$.

I now have an equation for ${Y_t}({s_t})$ in terms of $s_x$ and $s_t$.

1st question (now solved) is how I work out the inverse Laplace transform wrt to t $L^{-1}_t$ of functions like $s_x$ ?

2nd question (unanswered): If y is a function of x and t, does this mean that x is a function of t?

3rd question (unanswered / not clear): Why are the above differential equations PDEs not ODEs?

Updates since comments:

I made a mistake on my working (I took transforms of x also when it should have been t) - I have now got $Y_t$ in terms of $s_t$ only, and can hence solve.

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  • $\begingroup$ I'm new to mathjax. Cleaned it up as best as I can. Not sure how to condense with less breaks between mathjax scripts $\endgroup$ Jul 10, 2016 at 12:45
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    $\begingroup$ Try wrapping the tex expressions you want to remain on the same line with a single set of "\$" symbols rather than with a double set, i.e. "\$\$". For example "\$x\$" leaves $x$ on the same line. Whilst "\$\$x\$\$" creates a new line for it: $$x.$$ $\endgroup$
    – okrzysik
    Jul 10, 2016 at 13:01
  • $\begingroup$ Also, are you sure that these equations are ODEs and not PDEs? $\endgroup$
    – okrzysik
    Jul 10, 2016 at 13:06
  • $\begingroup$ They are ODEs. I thought up the question myself. $\endgroup$ Jul 10, 2016 at 13:13
  • $\begingroup$ I see. The problem appears not to make any sense in its current form. $\endgroup$
    – okrzysik
    Jul 10, 2016 at 13:17

1 Answer 1

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Let $y(x)=z(x)$ (otherwise, after the Laplace transform they look like the same functions):

$$ \begin{cases} 3\cdot\frac{\text{d}^2y(t)}{\text{d}t^2}+\frac{\text{d}z(x)}{\text{d}x}=0\\ 5\cdot\frac{\text{d}^2z(x)}{\text{d}x^2}-\frac{\text{d}y(t)}{\text{d}t}=0 \end{cases} $$

Take the Laplace transform of both sides:

-First ODE: $$3\cdot\left(s^2\text{Y}(s)-sy(0)-y'(0)\right)+\left(s\text{Z}(s)-z(0)\right)=0$$ -Second ODE: $$5\cdot\left(s^2\text{Z}(s)-sz(0)-z'(0)\right)-\left(s\text{Y}(s)-y(0)\right)=0$$

Now, simplify and factor out:

-First ODE: $$\text{Y}(s)=\frac{3sy(0)+3y'(0)+z(0)-s\text{Z}(s)}{3s^2}$$ -Second ODE: $$\text{Z}(s)=\frac{5sz(0)+5z'(0)+s\text{Y}(s)-y(0)}{5s^2}$$

Now, using substitution:

  • For $\text{Y}(s)$: $$\text{Y}(s)=\frac{y(0)+15s^2y(0)+15sy'(0)-5z'(0)}{s+15s^3}$$

  • For $\text{Z}(s)$: $$\text{Z}(s)=\frac{z(0)+15s^2z(0)+3y'(0)+15sz'(0)}{s+15s^3}$$

Now, with Inverse Laplace transform:

  • For $y(t)$: $$y(t)=y(0)+5z'(0)\left(\cos\left(\frac{t}{\sqrt{15}}\right)-1\right)+y'(0)\sqrt{15}\sin\left(\frac{t}{\sqrt{15}}\right)$$

  • For $z(x)$: $$z(x)=z(0)-3y'(0)\left(\cos\left(\frac{x}{\sqrt{15}}\right)-1\right)+z'(0)\sqrt{15}\sin\left(\frac{x}{\sqrt{15}}\right)$$

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