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I've encountered a problem where I have to solve a volterra integro-differential equation of the following format. I tried different approaches but the exponential term makes the life miserable. Any suggestions?
$-u'(t) = a^2\int_0^tdx u(x) \mathrm{e}^{-b^2(t-x)^2-c(t-x)}$
with the initial condition $u(t) = 1$. Thanks in advance!
$$-{\frac {\rm d}{{\rm d}t}}u \left( t \right) ={a}^{2}\int_{0}^{t}\!u \left( x \right) {{\rm e}^{-{b}^{2} \left( t-x \right) -c \left( t-x \right) }}\,{\rm d}x $$
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 $
