For $$x \in R$$, $$x \neq 0$$, if $$y(x)$$ is a differentiable function such that $$x\int_1^x y(t)dt = (x+1)\int_1^x ty(t)dt$$, then $$y(x)$$ equals (where C is a constant)

$$x \int_{1}^{x} y(t) dt = (x + 1) \int_{1}^{x} ty(t) dt$$

$$\frac{d}{dx} \left[ x \int_{1}^{x} y(t) dt \right] = \frac{d}{dx} \left[ (x + 1) \int_{1}^{x} ty(t) dt \right]$$

Applying the product rule and Leibniz rule:

$$1 \cdot \int_{1}^{x} y(t) dt + x \cdot y(x) = 1 \cdot \int_{1}^{x} ty(t) dt + (x + 1) \cdot x y(x)$$

$$\int_{1}^{x} y(t) dt + xy = \int_{1}^{x} ty(t) dt + x^2y + xy$$

$$\int_{1}^{x} y(t) dt = \int_{1}^{x} ty(t) dt + x^2y$$

$$\frac{d}{dx} \left[ \int_{1}^{x} y(t) dt \right] = \frac{d}{dx} \left[ \int_{1}^{x} ty(t) dt + x^2y \right]$$

$$y = xy + \left( 2xy + x^2 \frac{dy}{dx} \right)$$

$$y = 3xy + x^2 \frac{dy}{dx}$$

$$y - 3xy = x^2 \frac{dy}{dx}$$

$$\frac{dy}{y} = \frac{1 - 3x}{x^2} dx$$

$$\frac{dy}{y} = \left( \frac{1}{x^2} - \frac{3}{x} \right) dx$$

Integrating both sides: $$\int \frac{1}{y} dy = \int \left( x^{-2} - \frac{3}{x} \right) dx$$

$$\ln y = -\frac{1}{x} - 3 \ln x + \ln C$$

$$\ln y = \ln e^{-1/x} + \ln x^{-3} + \ln C$$

$$\ln y = \ln \left( \frac{C}{x^3} e^{-1/x} \right)$$

$$y(x) = \frac{C}{x^3} e^{-1/x}$$