Solution: Solve f(x)=x2+\int_0^x e^{-t}f(x-t)dt for f(x)

[himanshu121]

$$f(x)=x^2+\int_0^x e^{-t}f(x-t)dt$$ ...(I)

Find f(x)

Okay what i did:

For
$$\int_0^x e^{-t}f(x-t)dt$$

I substituted h=x-t =>dh=-dt
so $$\int_0^x e^{h-x}f(h)dh$$

Now i differentiated (I)
so i got
f'(x)=2x+f(x) after solvin this by integrating factor method i got different results which involve e^x

But i had to prove f(x)=x²+x³/3

 

[Hurkyl]

The fundamental theorem of calculus says

$$\frac{d}{dx} \int_a^x f(h) \, dh = f(x)$$

However, when you differentiate, you don't have something in this form! You have the form

$$\frac{d}{dx} \int_a^x f(\mathb{x,} h) \, dh$$

Fortunately, in this case, you don't need the full messy version; you can factor $e^{-x}$ out of the integrand and then differentiate normally.

 

[himanshu121]

Thanks Hurkyl I got corrected

now i have f'(x)=x²+2x

which gives me the result

Still i want to know how would u have approached the pro