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Bowei Tang
Bowei Tang
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Let $v=x-t,$ $$\int_{0}^{x}e^{-t}f(x-t)dt=e^x\int_{0}^{x}e^{-v}f(v)dv$$$$\int_{0}^{x}e^{-t}f(x-t)dt=e^{-x}\int_{0}^{x}e^{v}f(v)dv$$

Differentiate the equation on both sides w.r.t $x$ gives:$$f'(x)=2x+f(x)+e^x\int_{0}^{x}e^{-v}f(v)dv=2x+f(x)+f(x)-x^2$$$$f'(x)=2x+f(x)-e^{-x}\int_{0}^{x}e^{v}f(v)dv=2x+f(x)-f(x)+x^2$$ $$f'(x)-2f(x)=2x-x^2$$$$f'(x)=2x+x^2$$ Then solve the ODE with initial condition $f(0)=0$.

Let $v=x-t,$ $$\int_{0}^{x}e^{-t}f(x-t)dt=e^x\int_{0}^{x}e^{-v}f(v)dv$$

Differentiate the equation on both sides w.r.t $x$ gives:$$f'(x)=2x+f(x)+e^x\int_{0}^{x}e^{-v}f(v)dv=2x+f(x)+f(x)-x^2$$ $$f'(x)-2f(x)=2x-x^2$$ Then solve the ODE with initial condition $f(0)=0$.

Let $v=x-t,$ $$\int_{0}^{x}e^{-t}f(x-t)dt=e^{-x}\int_{0}^{x}e^{v}f(v)dv$$

Differentiate the equation on both sides w.r.t $x$ gives:$$f'(x)=2x+f(x)-e^{-x}\int_{0}^{x}e^{v}f(v)dv=2x+f(x)-f(x)+x^2$$ $$f'(x)=2x+x^2$$ Then solve the ODE with initial condition $f(0)=0$.

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Bowei Tang
Bowei Tang
  • 1.5k
  • 3
  • 18

Let $v=x-t,$ $$\int_{0}^{x}e^{-t}f(x-t)dt=e^x\int_{0}^{x}e^{-v}f(v)dv$$

Differentiate the equation on both sides w.r.t $x$ gives:$$f'(x)=2x+f(x)+e^x\int_{0}^{x}e^{-v}f(v)dv=2x+f(x)+f(x)-x^2$$ $$f'(x)-2f(x)=2x-x^2$$ Then solve the ODE with initial condition $f(0)=0$.