find a function such...

Question

exercise : A function $f$, continuous on the positive real axis, has the property that $$\int_{1}^{xy}f(t)dt =y\int_{1}^{x}f(t)dt +x\int_{1}^{y}f(t)dt$$ for all $x > 0$ and all $y > 0$. If $f (1) = 3$, compute $f (x)$ for each $x > 0$.

My progress: I derive $f(xy)y = yf(x)+\int_{1}^{y}f(t)dt$ then $x=1$ $$f(y)y=3y+\int_{1}^{y}f(t)dt$$ then don't know what to do.

score 3 · Answer 1 · answered Dec 08 '15 at 16:35

3

Derive once more to get $$f'(y)y+f(y)=3+f(y)$$ then you have (changing from $y$ to $x$) $$f'(x)x=3$$ $$f'(x)=\frac3x$$ $$f(x)=\int \frac3x\,dx = 3\ln(x)+C,\quad \text{for } x>0$$ Then you can find $C$: $$f(1)=3\implies 3\ln(1)+C=3\implies C=3$$ $$f(x)=3\ln(x)+3$$

answered Dec 08 '15 at 16:35

Kamil Jarosz

4,984

Thanks. I thought that if I kept deriving left me a differential equation. – Elll Dec 08 '15 at 16:45
@Elll actually, it is a differential equation, but a simple one :) – Kamil Jarosz Dec 08 '15 at 16:46

find a function such...

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