# Which inverse Laplace form can I use?

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1. Oct 17, 2016

### kostoglotov

1. The problem statement, all variables and given/known data

I have the second order diff eq:

Solving by Laplace transform gets me to:

I could use the inverse laplace transform that takes me back to $e^{at}cos(bt)$ with b=0, but that only solves for the homogeneous (complementary) part of the equation, it won't reproduce the dirac delta function.

What should I manipulate the Y(s) into? Should I tackle it with partial fraction decomposition?

2. Relevant equations

3. The attempt at a solution

2. Oct 17, 2016

### Staff: Mentor

$Y(s) =\frac{}{} = \frac{\alpha s}{(s - 4)^2} + \frac{\kappa - 8\alpha + \beta}{(s - 4)^2}$
Take the inverse Laplace transform of each of the two parts on the right. The numerator on the right is just a constant. A table of inverse Laplace transforms should have both of these.

3. Oct 18, 2016

### Ray Vickson

If you want to have $y(x) = 0$ for $x < 0$, then the DE + initial conditions are compatible only if $\alpha = 0$ and $\kappa = \beta$. To see this, look at the left-hand-side near $x=0$, and interpret the derivatives in terms of generalized functions ($\delta (x)$, etc.). If $\alpha \neq 0$ we have $y'(x) = \alpha \delta (x) + \; \text{smooth terms}$ and $y''(x) = \alpha \delta'(x) + \beta \delta (x) + \: \text{smooth terms}$, so we would need
$$\alpha \delta'(x) +\beta \delta(x) - 8 \alpha \delta (x) = \kappa \delta(x).$$
This fails because $\delta$ and $\delta'$ are very different as generalize functions. On the other hand, if $y(x)$ is continuous at $x = 0$, so that $\alpha = 0$, we have $\beta \delta (x) = \kappa \delta (x)$, and this works when $\kappa = \beta$.

If $\alpha, \beta$ and $\kappa$ are three different parameters, we cannot have $y(x) = 0$ for $x < 0$, and so we cannot solve the problem using Laplace transforms. We could try a Fourier transform instead, but by far the easiest way would be the direct approach where we apply different valuations of the homogeneous solution for $x > 0$ and for $x < 0$, then match up the boundary conditions at $x = 0$. By this I mean: use homogeneous solution $y(x) = a_1\, y_1(x) + a_2 \, y_2(x)$ for $x < 0$ and $y(x) = b_1\, y_1(x) + b_2 \,y_2(x)$ for $x > 0$ (with the same functions $y_1(x)$ and $y_2(x)$ in both). Assuming $y$ continuous at $x = 0$ we get $y(0-) = \alpha$ and $y(0+) = \alpha$. Next: integrate the DE from $x = -\epsilon$ to $x = +\epsilon$ and then look at the limit as $\epsilon \to 0$. That gives $y'(\epsilon)- y'(-\epsilon) + O(\epsilon) = \kappa$, so the condition is $y'(0+) - y'(0-) = \kappa$. Thus, $y'(0+) = \beta$ and $y'(0-) = \beta - \kappa$. Those four conditions allow determination of $a_1,a_2, b_1,b_2$.

Last edited: Oct 18, 2016