IE17

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Solve: u(x)=\int _{0}^{x}e^{{x-y}}u(y)dy,u(0)=0\,

u(x)=e^{x}\int _{0}^{x}e^{{-y}}u(y)dy\,

u'(x)={\frac  {d}{dx}}\left[e^{x}\int _{0}^{x}e^{{-y}}u(y)dy\right]\,

=e^{x}\int _{0}^{x}e^{{-y}}u(y)dy+e^{x}\left[{\frac  {d}{dx}}\int _{0}^{x}e^{{-y}}u(y)dy\right]\,

u'(x)=u(x)+e^{x}e^{{-x}}u(x)\,

u'(x)=2u(x)\,

u(x)=c_{1}e^{{2x}}\,

u(0)=0\implies c_{1}=0\,

Therefore

u(x)=0\,


Note: Infact any homogenous Volterra type equation gives a trivial solution.


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