Since you didn't respond to those pages I linked in comments, I'm guessing none of it makes any sense. So I'll keep it short and simple.
You know, or should know, that \$i_C=C\cdot\frac{\text{d}}{\text{d}t}v_x\$. You also know that the current in the resistor is obviously \$i_R=\frac{v_x}{R}\$. And the dependent current source is specified, as well. So you have (subtracting the resistor current from the dependent source current to get the capacitor current):
$$i_C=0.75\mho\cdot v_x-\frac{v_x}{4\:\Omega}=0.25\:\text{F}\cdot\frac{\text{d}}{\text{d}t}v_x$$
(The resistor just reduces the magnitude of the dependent source, is all. Technically, you could just disconnect the resistor after modifying the factor of 0.75 down to 0.5 on the source.)
So, just re-arrange to get:
$$\frac{0.5\mho}{0.25\:\text{F}}\cdot \text{d}t=\frac{\text{d}v_x}{v_x}$$
The solution is \$v_x=A\cdot\exp\left(2\cdot t\right)\$. But you know the initial condition, so the specific solution must be:
$$v_x=10\:\text{V}\cdot\exp\left(2\:\text{Hz}\cdot t\right)$$
In one second, \$v_x\approx 73.89\:\text{V}\$ about which LTspice appears to agree on:
Now, if none of this makes sense then there's some serious remedial work ahead. I like the book "Fundamentals of Differential Equations" by R. Kent Nagle & Edward B. Saff & Arthur David Snider and would recommend it for self-learners as an adjacent line of inquiry.
$v_c(0^{-})=10V$
should change to\$v_c(0^{-})=10V\$
i.e. \$v_c(0^{-})=10V\$ (if that's what you meant to write). \$\endgroup\$