What is i(t) from circuit with differential equation?
The proper way to form the homogeneous solution is (not what you have given in the fourth equation):
$${i_L}_h=C_1*e^{(-\frac{1}{2})t}$$
(Your differential equation is written for \$i_L(t)\$, it is better not to change it to \$y\$ midstream.)

For t>0, without knowing \$J\$, you cannot solve \$i_L(t)\$. The only specification of \$J\$ given is that it is 4. So a forced assumption is that \$J = 4\$ for all time if you want a solution from what are given.

With \$J\$ being constant, by inspection, $$i_L(t) = \frac{J_t}{5} $$ is a particular solution to the differential equation. You can plug that into your second equation to confirm.

Now the complete solution becomes:
$$i_L(t)=C_1*e^{(-\frac{1}{2})t} + \frac{J_t}{5}$$

Now you need the initial condition of \$i_L(0)\$ to solve for the constant \$C_1\$. Given \$J\$ is constant and assuming steady state at t=0, \$L\$ behaves as a short circuit at t=0. Therefore, the only elements left in the schematic for figuring out the initial condition are \$J,\ R_2,\ R_3\$, a simple current divider.