In that case, ##y## is ##\theta## and the rest is the same, although I think Wolfram Alpha must have had a factor of ##t## on the ##c_1^2##, did it not?
Since ##\alpha = d\omega / dt##, you can rewrite the first equation as:
$$\frac{d\omega}{ dt} = 50\sqrt{\omega}$$
And then through separation of variables:
$$\frac{d\omega}{50\sqrt{\omega}} = dt$$
Then integrate both sides:
$$\frac{1}{50}\int\frac{d\omega}{\sqrt{\omega}} = \int dt$$
$$\frac{1}{25}\sqrt{\omega} = t + c_1$$
$$\omega= 625t^2 + 1250c_1t + 625c_1^2$$
Note that if we roll a factor of 25 into ##c_1##, this becomes:
$$\omega= 625t^2 + 50c_1t + c_1^2$$
This is ok to do because the constants are arbitrary. And since ##\omega = d\theta / dt##, you can repeat the process:
$$\frac{d\theta}{dt} = 625t^2 + 50c_1t + c_1^2$$
$$d\theta = (625t^2 + 50c_1t + c_1^2) dt$$
$$\int d\theta = \int (625t^2 + 50c_1t + c_1^2) dt$$
$$\theta = \frac{625}{3}t^3 + 25c_1t^2 + c_1^2 t + c_2$$
As for the boundary conditions, you can redo the above using the boundary conditions as the limits of integration. Or, you can use the above equation to plug the values for ##y## and ##t## that you are given, and solve for the arbitrary constants. You'll need two conditions to solve for both constants.