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Solve this integral equation using Laplace transform

$$f(x)=x^2 + \int_{0}^{x}f^{\prime}(x-t) e^{-at} dt ,f(0)=0 $$

Please Help

see mu answer below

Thank you for your participation

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My Solution:-

Applying Laplace transform , we get

$$\bar f(s)=\frac{2!}{s^3}+L[f^{\prime}(x);s].L[e^{-at};s]$$ $$\bar f(s)=\frac{2}{s^3}+[s\bar f(s)-f(0)].[\frac{1}{s+a}]$$ $$\bar f(s)=\frac{2}{s^3}+\frac{s\bar f(s)}{s+a}$$ $$(1-\frac{s}{s+a})\bar f(s)=\frac{2}{s^3}$$ $$(\frac{a}{s+a})\bar f(s)=\frac{2}{s^3}$$ $$\bar f(s)=\frac{2(s+a)}{as^3}$$ $$\bar f(s)=\frac{2}{a}[\frac{1}{s^2}+\frac{a}{s^3}]$$ Applying the inverse transform , we get

$$f(x)=\frac{2}{a}[x+\frac{a}{2}x^2]$$ $$f(x)=\frac{2}{a}x+x^2$$

Is this true solution?

Is there a simplification of the final answer

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