Practicing Success
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). |
$C x^3 e^{1 / x}$ $\frac{C}{x} e^{-1 / x}$ $\frac{C}{x^2} e^{-1 / x}$ $\frac{C}{x^3} e^{-1 / x}$ |
$\frac{C}{x^3} e^{-1 / x}$ |
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 $-\frac{1}{x}-3 \log x=\log y(x)+\log K$ $\Rightarrow -\frac{1}{x}=\log \left(K x^3 y(x)\right)$ $\Rightarrow K x^3 y(x)=e^{-1 / x} \Rightarrow y(x)=\frac{1}{K x^3} e^{-1 / x} \Rightarrow y(x)=\frac{C}{x^3} e^{-1 / x}$ |