For $x \in R, x \neq 0$, if $y(x)$ is differentiable function such that $x \int\limits_1^x y(t) d t=(x+1) \int\limits_1^x t y(t) d t$, then $y(x)$ equals (where C is a constant).
Answer & explanation
Correct answer: option 4
We have,
$x \int\limits_1^x y(t) d t=(x+1) \int\limits_1^x t y(t) d t$ .....(i)
Differentiating with respect to $x$, we get
$\int\limits_1^x y(t) d t+x y(x)=\int\limits_1^x t y(t) d t+(x+1) x y(x)$
$\Rightarrow \int\limits_1^x y(t) d t=\int\limits_1^x t y(t) d t+x^2 y(x)$
Differentiating with respect to $x$, we get
$y(x)=x y(x)+x^2 \frac{d y}{d x}(x)+2 x y(x)$
$\Rightarrow (1-3 x) y(x)=x^2 \frac{d}{d x}(y(x))$
$\Rightarrow \left(\frac{1}{x^2}-\frac{3}{x}\right) d x=\frac{d(y(x))}{y(x)}$
Integrating, we obtain
$-\frac{1}{x}-3 \log x=\log y(x)+\log K$
$\Rightarrow -\frac{1}{x}=\log \left(K x^3 y(x)\right)$
$\Rightarrow K x^3 y(x)=e^{-1 / x} \Rightarrow y(x)=\frac{1}{K x^3} e^{-1 / x} \Rightarrow y(x)=\frac{C}{x^3} e^{-1 / x}$