= Solution
Let $J=I_1=I_2$, $\omega_3=c$ and
$$
z(t)=\omega_1(t)+i\omega_2(t),\qquad
\nu=\frac{(I_3-J)c}{J}.
$$
Here $c$ and $\nu$ are constant. The first two of <Euler's equations (rigid body dynamics)> give $\dot\omega_1=-\nu\omega_2+\tau_0/J$ and $\dot\omega_2=\nu\omega_1+\tau_0/J$. Combining them produces
$$
\boxed{\dot z=i\nu z+\frac{\tau_0}{J}(1+i).}
$$
Multiplication by the <integrating factor> $e^{-i\nu t}$ gives, for $\nu\ne0$,
$$
z(t)=z(0)e^{i\nu t}+\frac{\tau_0(1+i)}{i\nu J}\left(e^{i\nu t}-1\right).
$$
For completeness, $\nu=0$ instead gives $z(t)=z(0)+\tau_0(1+i)t/J$. These formulas describe a <symmetric top under constant transverse body torque>.
The stated <initial conditions> require $b\ne0$ and $I_3\ne J$. They give $c=b/(I_3-J)$, $\nu=b/J$ and $z(0)=i\tau_0/b$. Substitution and simplification yield
$$
z(t)=\frac{\tau_0}{b}\left(e^{ibt/J}-1+i\right).
$$
Taking the <real part> and <imaginary part>, the <angular velocity> in the <principal body frame> is
$$
\boxed{\boldsymbol\omega(t)=\left(
\frac{\tau_0}{b}\left[\cos\left(\frac{bt}{I_1}\right)-1\right],\quad
\frac{\tau_0}{b}\left[\sin\left(\frac{bt}{I_1}\right)+1\right],\quad
\frac{b}{I_3-I_1}\right).}
$$
Its value at $t=0$ is the required <initial condition>; differentiating the first two components verifies both forced <Euler's equations (rigid body dynamics)>.
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