$$-{\frac {\rm d}{{\rm d}t}}u \left( t \right) ={a}^{2}\int_{0}^{t}\!u
\left( x \right) {{\rm e}^{-{b}^{2} \left( t-x \right) -c \left( t-x
\right) }}\,{\rm d}x
$$
Using Laplace transform:
$$-s*{\it laplace} \left( u \left( t \right) ,t,s \right) +u \left( 0
\right) ={\frac {{a}^{2}{\it laplace} \left( u \left( t \right) ,t,s
\right) }{{b}^{2}+c+s}}
$$
$${\it laplace} \left( u \left( t \right) ,t,s \right) ={\frac {u
\left( 0 \right) \left( {b}^{2}+c+s \right) }{{b}^{2}s+{a}^{2}+cs+{s
}^{2}}}
$$
then inverse:
$u \left( t \right) ={{\rm e}^{-{\frac {t \left( {b}^{2}+c \right) }{2}
}}}u \left( 0 \right) \left( \cosh \left( {\frac {t}{2}\sqrt {
\left( {b}^{2}+c \right) ^{2}-4\,{a}^{2}}} \right) +{({b}^{2}+c)\sinh
\left( {\frac {t}{2}\sqrt { \left( {b}^{2}+c \right) ^{2}-4\,{a}^{2}}
} \right) {\frac {1}{\sqrt { \left( {b}^{2}+c \right) ^{2}-4\,{a}^{2}}
}}} \right)$
where: $ u(0) =1 $