If I define $f(t)=\int_0^t y(\tau)d\tau$, I need to solve
$$f''(t)=\alpha(t)+f(t)^\gamma.$$
For $\alpha\equiv 0$ this has the implicit solution

$${({\gamma}+1) f(t)^2 \left(c_1 {\gamma}+c_1+2 f(t)^{{\gamma}+1}\right)^2 \, _2F_1\left(\frac{1}{2},\frac{1}{{\gamma}+1};1+\frac{1}{{\gamma}+1};-\frac{2 f(t)^{{\gamma}+1}}{{\gamma} c_1+c_1}\right)^2}{}=\left(c_2+t\right)^2\,\left(c_1 {\gamma}+c_1\right) \left(c_1 ({\gamma}+1)+2 f(t)^{{\gamma}+1}\right)^2$$

Closed-form expressions in terms of special functions (Weierstrass $\wp$ and Jacobi elliptic function sn, courtesy of Mathematica) are possible for small integer $\gamma$:

$$f(t)=c_1 e^t+c_2 e^{-t},\;\;\gamma=1$$
$$f(t)=\sqrt[3]{6}\; \wp \left(\frac{t+c_1}{\sqrt[3]{6}};0,c_2\right),\;\;\gamma=2$$
$$f(t)=\pm\sqrt[4]{2} \sqrt{ic_1}\; \text{sn}\left(\left.\frac{(-1)^{3/4} \sqrt{\sqrt{2} \sqrt{c_1} t^2+2 c_2 \sqrt{2} \sqrt{c_1} t+c_2^2 \sqrt{2} \sqrt{c_1}}}{\sqrt{2}}\right|-1\right),\;\;\gamma=3$$

More generally, for constant $\alpha$, the implicit solution is
$$\int_1^{f(t)} \left({2\alpha x+\frac{2}{\gamma+1}x^{\gamma+1}+c_1}\right)^{-1/2} \, dx=c_2+t$$
with explicit solutions

$$f(t)=-\alpha+c_1 e^t+c_2 e^{-t},\;\;\gamma=1$$
$$f(t)=\sqrt[3]{6} \;\wp \left(\frac{t+c_1}{\sqrt[3]{6}};-2 \sqrt[3]{6} \alpha,c_2\right),\;\;\gamma=2$$