Let f be a differentiable function such that x^2f(x)-x=4 _0^xtf(t)dt, f(1)=(2)/(3).

Question

Let $f$ be a differentiable function such that ${x}^{2}f(x)-x=4\int _{0}^{x}tf(t)dt$, $f(1)=\frac{2}{3}$. Then $18f(3)$ is equal to :

Accepted Answer

Given that

$$
x^2 f(x)-x=4 \int_0^x t f(t) d t
$$

On differentiating both sides with respect to $x$, we get

$$
\[\begin{gathered} 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{gathered}$$
\]

On comparing above equation with

$\frac{d y}{d x}+P y=Q$, where $P=-\frac{2}{x}, Q=\frac{1}{x^2}$

Now, IF $=e^{\int(-2 / x) d x}=e^{-2 \log x}=1 / x^2$

Solution is $y\left(\frac{1}{x^2}\right)=\int\left(\frac{1}{x^2}\right) \frac{1}{x^2} d x$

$$
\[\begin{gathered} \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{gathered}$$
\]

Given, $f(1)=\frac{2}{3}$.

So, $\frac{2}{3}=-\frac{1}{3}+C \Rightarrow C=1$

Thus, $y=f(x)=-\frac{1}{3 x}+x^2$

$$
\therefore 18 f(3)=18\left\{9-\frac{1}{9}\right\}=18 \times \frac{80}{9}=160
$$

Solutions · 1