Let $f$ be a differentiable function such that ${x^2}f(x) - x = 4\int\limits_0^x {tf(t)dt}$, $f(1) = {2 \over 3}$. Then $18f(3)$ is equal to :

Given that <br/><br/>$x^2 f(x)-x=4 \int_0^x t f(t) d t$ <br/><br/>On differentiating both sides with respect to $x$, we get <br/><br/>$$ \begin{array}{rlrl} & 2 x f(x)+x^2 f^{\prime}(x)-1 =4 x f(x) \\\\ &\Rightarrow x^2 f^{\prime}(x)-2 x f(x)-1 =0 \\\\ &\Rightarrow x^2 \frac{d y}{d x}-2 x y =1 ~~~~~~~(Let, y=f(x) ]\\\\ &\frac{d y}{d x}-\frac{2}{x} y =\frac{1}{x^2} \end{array} $$ <br/><br/>On comparing above equation with <br/><br/>$\frac{d y}{d x}+P y=Q$, where $P=-\frac{2}{x}, Q=\frac{1}{x^2}$ <br/><br/>Now, IF $=e^{\int(-2 / x) d x}=e^{-2 \log x}=1 / x^2$ <br/><br/>Solution is $y\left(\frac{1}{x^2}\right)=\int\left(\frac{1}{x^2}\right) \frac{1}{x^2} d x$ <br/><br/>$$ \begin{array}{ll} &\Rightarrow \frac{y}{x^2}=\int \frac{1}{x^4} d x \Rightarrow \frac{y}{x^2}=-\frac{1}{3 x^3}+C \\\\ &\Rightarrow y =-\frac{1}{3 x}+C x^2 \end{array} $$ <br/><br/>Given, $f(1)=\frac{2}{3}$. <br/><br/>So, $\frac{2}{3}=-\frac{1}{3}+C \Rightarrow C=1$ <br/><br/>Thus, $y=f(x)=-\frac{1}{3 x}+x^2$ <br/><br/>$\therefore 18 f(3)=18\left\{9-\frac{1}{9}\right\}=18 \times \frac{80}{9}=160$