Question

Let $f:[1,\infty )\to [2,\infty )$ be a differentiable function such that $f\left(1\right)=2.$ If $6{\int }_1^{x}f\left(t\right)dt=3xf\left(x\right)-x^3\text{for all}x\ge 1,$ then the value of $f\left(2\right)$ is

Answer 1

From the given condition, we have
$6{\int }_1^{x}f\left(t\right)dt=3xf\left(x\right)-x^3$
$\Rightarrow 6f\left(x\right)=3f\left(x\right)+3x{f}^{\prime }\left(x\right)-3x^2$
$\Rightarrow x{f}^{\prime }\left(x\right)-f\left(x\right)=x^2$
$\Rightarrow f^{\prime }\left(x\right)-\frac{1}{x}f\left(x\right)=x$
which is linear differential equation, whose I.F.$=e^{-\int \frac{1}{x}dx}=e^{-\ln x}=\frac{1}{x}$
Solution of the given equation is
$f\left(x\right).\frac{1}{x}=\int x.\frac{1}{x}dx+c$
$\Rightarrow f\left(x\right)=x^2+cx$
Initially, for $x=1,f\left(1\right)=2\Rightarrow c=1$
$\Rightarrow f\left(x\right)=x^2+x$
$\Rightarrow f\left(2\right)=4+2=6$