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


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.