$\alpha$ is just a constant and it's given that
$$\frac{\partial Y(K,L)}{\partial K}=\alpha\frac{Y}{K}$$ $$\frac{\partial Y(K,L)}{\partial L}=(1-\alpha)\frac{Y}{L}$$
Doing some integration
$$\ln(Y)=\alpha \ln(K)+C_1$$ $$\ln(Y)=(1-\alpha) \ln(L)+C_2$$
Then
$$Y=C_3K^\alpha $$ $$Y=C_4L^{1-\alpha} $$
Since $Y$ is proportional to $K^\alpha$ and to $L^{1-\alpha}$ when the other variable is held constant, I'm very tempted to say that it's proportional to both, making
$$Y(K,L)=A K^\alpha L^{1-\alpha}$$
However, I couldn't find a way to prove that that's actually true
Your $C_3$ should depend on $L$ and $C_4$ on $K$, namely $$ Y=C_3(L)K^\alpha $$ and $$ Y=C_4(K)L^{1-\alpha}. $$ So $$ C_3(L)K^\alpha=C_4(K)L^{1-\alpha} $$ or $$ \frac{C_3(L)}{L^{1-\alpha}}=\frac{C_4(K)}{K^\alpha}\equiv A.$$ From this, one has $$ C_3(L)=AL^{1-\alpha},C_4(k)=AK^\alpha$$ and hence $$ Y=AL^{1-\alpha}K^\alpha. $$