# Why do these 2 methods give different solutions?

I need to solve what is underlined in red for $x_i$, nut currently I'm interested in the right side of the equation only.

On the left I sarted by doing the Laplace transform of $x_u'$ and $x_u$, and this gives the correct solution.

On the right side I tried to derive $x_u$ first, plug everything into the equation and then do the Laplace transform, and this gives a completely different solution. Is this method incorrect? Or am I doing something wrong?

The problem is that you took the derivative of the function

$$\hat{x}_u(t)=2e^{-3t}-e^{-4t}\tag{1}$$

whereas using the Laplace transform you implicitly assumed that $x_u(t)$ equals zero for $t<0$:

$$x_u(t)=\hat{x}_u(t)u(t)=(2e^{-3t}-e^{-4t})u(t)\tag{2}$$

where $u(t)$ is the unit step function.

If you take the derivative of $(2)$ then you'll get the same result as with the Laplace transform, taking into account that the derivative of the step function $u(t)$ is the Dirac delta impulse $\delta(t)$:

$$\frac{dx_u(t)}{dt}=\frac{d\hat{x}_u(t)}{dt}u(t)+\hat{x}_u(t)\delta(t)=\frac{d\hat{x}_u(t)}{dt}u(t)+\hat{x}_u(0)\delta(t)\tag{3}$$

where I've used the fact that for any function $f(t)$ that is continuous at $t=0$ we have $f(t)\delta(t)=f(0)\delta(t)$. I trust that you can take it from here.