Question

Let f: $[1, \infty) \rightarrow[2, \infty)$ be a differentiable function such that $f(1)=2$. If $6 \int\limits_{1}^{x} f(t) d t=3 x f(x)-x^{3}$ for all $x \geq$ 1, then the value of $f (2)$ is____

Answer 1

$6 \int\limits_{1}^{x} f(t) d t=3 x f(x)-x^{3}$
$\Rightarrow 6 f(x)=3 f(x)+3 x f'(x)-3 x^{2}$
$\Rightarrow 3 f(x)=3 x f'(x)-3 x^{2}$
$\Rightarrow x f'(x)-f(x)=x^{2}$
$\Rightarrow x \frac{d y}{d x}-y=x^{2}$
$\Rightarrow \frac{d y}{d x}-\frac{1}{x} y=x$ ...(i)
$I . F=e^{\int \frac{1}{x} d x}=e^{-\log _{e} x}$
Multiplying (i) both sides by $\frac{1}{x}$
$\frac{1}{x} \frac{d y}{d x}-\frac{1}{x^{2}} y=1$
$\Rightarrow \frac{d}{d x}\left(y \cdot \frac{1}{x}\right)=1$
integrating
$\frac{y}{x}=x +c$
Put $x=1, y=2$
$\Rightarrow 2=1+c$
$\Rightarrow c=1$
$\Rightarrow y=x^{2}+x$
$\Rightarrow f(x)=x^{2}+x$
$\Rightarrow f(2)=6$
Note: If we put $x=1$ in the given equation we get $f(1)=1 / 3$.