Help solving a 2nd-order RLC circuit
The differential equation:
Current \$i\$ flows through the right branch and \$3-i\$ flows through left branch. So we can write:

$$10(3-i)+20\int(3-i)dt = 4i+2\frac{di}{dt}\tag1$$ 

which can be differentiated and re-arranged to get the equation:

$$y'' + 7y' + 10y = 30\tag2$$

The equation you obtained assumes that same current \$i\$ flows through both the branches. 

The initial conditions:
The first initial condition you obtained is correct but the second one is wrong.

$$\frac{di}{dt}(0)\ne 0$$

At \$t=0,\$ capacitor can be considered as short circuit and inductor can be considered as open circuit. Then voltage across current source and hence the voltage across inductor will be 30V.

$$2\frac{di}{dt}(0) = 30$$
$$\Rightarrow \frac{di}{dt}(0) = 15$$

so the initial conditions are:

$$i(0) = 0; \frac{di}{dt}(0) = 15\tag3$$

Solving the differential equation in (2) with initial conditions in (3) will give you the required answer.