Integrate twice the differential equation $$ \frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}=\frac{w}{2}\left(l x-x^{2}\right) $$ where \(w\) and \(l\) are constants, to find a general solution for \(y\).

Integrate the second-order differential equation with respect to x: $$ \int\frac{d^2 y}{dx^2}dx = \int w \left(\frac{1}{2}(lx-x^2)\right)dx $$ Let \(v=\frac{dy}{dx}\), then by integrating both sides, we have: $$ \int dv = \int w \left(\frac{1}{2}(lx-x^2)\right)dx $$

When we integrate the right side of the equation, we get: $$ v = \int w \left(\frac{1}{2}(lx-x^2)\right)dx = \frac{1}{2}w\left(\frac{lx^2}{2}-\frac{x^3}{3}\right) + C_1 $$ Now, let’s integrate again to find y: $$ y = \int \left(\frac{1}{2}w\left(\frac{lx^2}{2}-\frac{x^3}{3}\right) + C_1\right)dx $$ Integrating the above equation, we obtain the general solution for y: $$ y = \frac{1}{2}w\left(\frac{lx^3}{6}-\frac{x^4}{12}\right) + C_1x + C_2 $$ Thus, the general solution for the given differential equation is: $$ y(x) = \frac{1}{2}w\left(\frac{lx^3}{6}-\frac{x^4}{12}\right) + C_1x + C_2 $$ Here, \(w\) and \(l\) are constants, and \(C_1\) and \(C_2\) are the integration constants representing, respectively, the initial slope and displacement of the function.