circuit analysis transient response in an RC circuit
You don't need to write down any differential equation to solve this circuit. Just use the fact that, assuming the system starts from rest, at the moment current source is connected to the circuit (at say t = 0), the capacitors are going to behave as short circuits and in DC steady state the capacitors will act as open circuits.

At \$t = 0\$

With \$C_1\$ &amp; \$C_2\$ replaced by short circuit, the equivalent circuit becomes just the parallel combination of \$R_2\$ and \$R_4\$. Giving the following currents in each branch:
$$I_1(0) = \frac{R_4I}{R_2+R_4}$$
$$I_2(0) = \frac{R_2I}{R_2+R_4}$$

At \$t = \infty\$

With \$C_1\$ &amp; \$C_2\$ replaced by open circuit, the equivalent circuit has parallel combination of \$R_1+R_2\$ and \$R_3 + R_4\$. Current in each branch:
$$I_1(\infty) = \frac{(R_3+R_4)I}{R_1 + R_2 + R_3 +R_4}$$
$$I_2(\infty) = \frac{(R_1+R_2)I}{R_1 + R_2 + R_3 +R_4}$$

The current \$I_1\$ in the above branch starts from \$I_1(0)\$ and reaches \$I_1(\infty)\$ with some time constant \$\tau_1\$, thus the current equation with time should be given by:
$$I_1(t) = I_1(\infty) + (I_1(0) - I_1(\infty))e^\frac{-t}{\tau_1}$$
The equivalent resistance seen by \$C_1\$ can be shown to be:
$$R_{C1eq} = \frac{R_1(R_2+R_3+R_4)}{R_1 + R_2 + R_3 +R_4}$$
And, \$\tau_1 = R_{eqC1}C_1\$
Similarly, \$I_2(t)\$ can be found.
Of course, you can also use Laplace transform to get the result, but this approach is easier and more intuitive.
It can be easily shown why the capacitors behave as open or short circuit as explained above.
The I-V characteristic of  a capacitor is given by:
$$i = C\frac{dv}{dt}$$ 
Clearly, at DC steady state, the voltage does not change and the current is zero,  resembling an open circuit.
Rearranging the equation, we can understand why it is a short circuit for initial period:
$$dv = \frac{idt}{C}$$
For small change in time, the change in voltage will be negligible. Since the capacitor is uncharged in the beginning implying zero volts across its terminals, voltage drop across it is going to be zero around this time. And zero volts implies short circuit.