 Solution

Write $t=T-T_c$, take ${\alpha }_2=at$ with $a>0$, and include the magnetic term $-Bm$. At $B=0$, minimizing the Landau free energy gives $m=0$ for $t>0$ andfor $t<0$, so the order-parameter critical exponent is $\beta =1/2$. Substitution gives a singular free-energy density proportional to $-t^2$ below $T_c$, whose second temperature derivative has a finite jump, so the heat-capacity exponent is $\alpha =0$. Above $T_c$, the equation ${\alpha }_{2}m+{\alpha }_4{m}^3=B$ gives the magnetic susceptibility $\chi =(\partial m/\partial B{)}_{B=0}=1/{\alpha }_2\sim t^{-1}$, hence $\gamma =1$. At $t=0$, $B={\alpha }_4{m}^3$, so $m\sim B^{1/3}$ and $\delta =3$.