From transfer function to differential equation
Given a transfer function
$$G_v(s) = \frac{k_v}{1 + sT} \tag{1}$$
the corresponding LCCDE, with $y(t)$ being the solution, and $x(t)$ being the input, will be
$$ T ~\dot{y}(t) + y(t) = k_v ~x(t) \tag{2} $$
Your formulation replaces $x(t)$ with a unit-step $u(t)$, and $y(t)$ with $x(t)$, yielding
$$ T ~\dot{x}(t) + x(t) = k_v ~u(t) \tag{3} $$
or equivalently
$$ \dot{x}(t) + \frac{1}{T} x(t) = \frac{k_v}{T} ~u(t) \tag{4} $$
Eq.4 represents a typical first order, constant coefficient, linear, ordinary differential equation (abbr LCCDE) whose solution procedure is as follows:
First, find the homogeneous solution to the Eq.4 with RHS being zero, as
$$ x_h(t) = A e^{-t/T} \tag{5} $$
where $A$ is an unknown coefficient to be determined by the initial condition $x(0)$.
Second, find the particular solution of Eq.4 with the RHS being $u(t)$. For this you can use the method of undetermined coefficients, by assuming a solution of the type
$$ x_p(t) = B u(t) \tag{6}$$
an letting $t \to \infty$, will yield in Eq.4 as
$$ 0 + \frac{B}{T} = \frac{k_v}{T} \tag{7} $$
where you find $B$ as
$$ B = k_v \tag{8}$$
Then you find the complete solution by adding the homogeneous part to the particular part, and solving for the unknown constant $A$ from the initial condition $x(0)$
$$ x(t) = x_h(t) + x_p(t) = A e^{-t/T} + k_v ~~~~,~~~~ t &gt;  0\tag{9} $$
The initial condition $x(0)$ is given as $0$ in the problem, hence
$$ x(0) = 0 = A + k_v \implies   A = -k_v \tag{10} $$
And the overall solution of the initial value problem in Eq.4 with $x(0)=0$ becomes:
$$ x(t) = -k_v e^{-t/T} + k_v = k_v (1 - e^{-t/T}) u(t) \tag{11} $$
Note that we look for a causal solution, unless otherwise stated, hence assume $ t&gt;0$.