= Solution
Treat the left side as a causal <convolution> with $t^{-\alpha}$. Under the usual Laplace-transform hypotheses, the <convolution theorem for Laplace transforms> gives
$$
\widehat\Phi(s)\,\Gamma(1-\alpha)s^{\alpha-1}=\widehat f(s).
$$
For an absolutely continuous $f$ with $f(0)=0$, the <derivative> transform is $\mathcal L[f']=s\widehat f$. Consequently
$$
\widehat\Phi(s)=\frac{s^{-\alpha}}{\Gamma(1-\alpha)}\mathcal L[f'](s).
$$
Since $\mathcal L[t^{\alpha-1}]=\Gamma(\alpha)s^{-\alpha}$, inverting the product proves
$$
\boxed{\Phi(t)=\frac1{\Gamma(\alpha)\Gamma(1-\alpha)}\int_0^tf'(\tau)(t-\tau)^{\alpha-1}d\tau.}
$$
Here $0<\alpha<1$ makes both power kernels locally integrable. For example assume exponential growth bounds sufficient for the transforms and a regularity class where the <convolution> is defined. This is the <Abel integral equation of fractional order>; the initial value removes the extra boundary term in the <derivative> transform.
For $f(t)=t$, the remaining integral is $\int_0^t(t-\tau)^{\alpha-1}d\tau=t^\alpha/\alpha$. Thus
$$
\boxed{\Phi(t)=\frac{t^\alpha}{\Gamma(\alpha+1)\Gamma(1-\alpha)}.}
$$
Substitution into the original equation checks the result: scaling $\tau=tu$ gives $t\,B(\alpha+1,1-\alpha)/[\Gamma(\alpha+1)\Gamma(1-\alpha)]=t$ by the <beta-gamma identity>.
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