Let f:RR be such that f(xy)=f(x)f(y) for all x,yR and f(0) 0.

Question

Let $f:\mathbb{R}\to\mathbb{R}$ be such that $f(xy)=f(x)f(y)$ for all $x,y\in\mathbb{R}$ and $f(0)\ne 0$. Let $g:[1,\infty)\to\mathbb{R}$ be a differentiable function such that
$$x^2g(x) = \int_1^x\big(t^2f(t)-tg(t)\big)dt.$$
Then $g(2)$ is equal to:

Accepted Answer

Setting $x=0$ in $f(xy)=f(x)f(y)$: $f(0)=f(0)f(y)$; since $f(0)\ne0$, $f(y)=1$ for all $y$. So $f(t)=1$.

The equation becomes $x^2g(x)=\int_1^x(t^2-tg(t))\,dt$.

Differentiate: $x^2g'(x)+2xg(x) = x^2-xg(x) \implies x^2g'(x)+3xg(x)=x^2 \implies g'(x)+\dfrac{3}{x}g(x)=1$.

I.F. $=x^3$. $\dfrac{d}{dx}(x^3g(x))=x^3 \implies x^3g(x) = \dfrac{x^4}{4}+C \implies g(x)=\dfrac{x}{4}+\dfrac{C}{x^3}$.

At $x=1$: original equation gives $g(1)=0$ (since RHS integral from 1 to 1 is 0). So $\dfrac14+C=0 \implies C=-\dfrac14$.

$g(x)=\dfrac{x}{4}-\dfrac{1}{4x^3}$. So
$$g(2) = \frac24-\frac{1}{32} = \frac12-\frac{1}{32} = \frac{15}{32}.$$

Solutions · 1