IEVS1

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du(x)=f(x)+\int _{0}^{x}k(x-y)u(y)\,dy

The integral is of convolution type. Take the Laplace transform of both sides.

{\mathcal  {L}}u={\mathcal  {L}}f+{\mathcal  {L}}k{\mathcal  {L}}u\,

{\mathcal  {L}}u={\frac  {{\mathcal  {L}}f}{1-{\mathcal  {L}}k}}\,

So the answer is the inverse Laplace transform of the right hand side.

u(x)={\mathcal  {L}}^{{-1}}\left({\frac  {{\mathcal  {L}}f}{1-{\mathcal  {L}}k}}\right)\,

Integral Equations

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