My problem is to solve
$$
\frac{\mathrm{d}^{2}x}{\mathrm{d}{t}^2} + 5\,\frac{dx}{dt} + 6x =
\delta\left(t\right)\quad\mbox{with}\quad x\left(0\right) = 0\quad
\mbox{and}\quad
\left.\frac{\mathrm{d}x}{\mathrm{d}{t}}\right\vert_{\ t\ =\ 0} = 0
$$
Taking the Laplace transform of both sides, I get
$(s^2+5s+6)L[x(t)]=L[\delta(t)]=1$
$L[x(t)]=\frac{1}{(s+2)(s+3)}=L[e^{-2t}-e^{-3t}]$
$x(t)=e^{-2t}-e^{-3t}$.
But this does not satisfy the equation or initial conditions
Could anyone help please? Thank you
Best Answer
Your solution is $x(t) = (e^{-2t}-e^{-3t})1_{t > 0}$ that is correct for the initial condition $x(t) = x'(t) = 0 $ for $t < 0$.
Note that the product rule (in the sense of distributions) applies, because it is the product of a distribution ($1_{t > 0}$) and a $C^\infty$ function, i.e. infinitely differentiable, ($e^{-at}$) : $$\frac{d}{dt} [e^{-at}1_{t > 0}] = 1_{t > 0}\frac{d}{dt} e^{-at}+e^{-at}\frac{d}{dt} 1_{t > 0} = - e^{-at}1_{t > 0}+ e^{-at}\delta(t) = - e^{-at}1_{t > 0}+ \delta(t)$$ hence $$x'(t) = -(2e^{-2t}-3e^{-3t})1_{t > 0}, \qquad x''(t) = (4e^{-2t}-9e^{-3t})1_{t > 0} +\delta(t)$$ and you can check that $x''(t)+5x'(t)+6x(t)= \delta(t)$.
Now since it is a linear differential equation, all the other solutions are of the form $\tilde{x}(t) = x(t)+y(t)$ where $y(t)$ is a solution of the homogeneous equation $y''(t)+5y'(t)+6y(t)= 0$. The general solution of this being $y(t) = Ae^{-2t}+Be^{-3t}$ you get that the general solution of $\tilde{x}''(t)+5\tilde{x}'(t)+6\tilde{x}(t)= \delta(t)$ is $$\tilde{x}(t) = (e^{-2t}-e^{-3t})1_{t > 0}+Ae^{-2t}+Be^{-3t}$$ and none of them is $C^1$ at $t=0$, hence as stated your exercice is incorrect.