A curve such that the area of the trapezoid formed by coordinate axes, ordinate at an arbitrary point and the tangent at this point equals half the square of its abscissa is 

The tangent at any point $(x, y)$ meets the Y-axis at $\left(0,y-x\frac{dy}{dx}\right)$ . According to the given condition .

$\frac{1}{2}{x}^2=$Area of the trapezoid

$=\left(y+y-x\frac{dy}{dx}\right)x$ 

$\Rightarrow \frac{dy}{dx}-2\frac{y}{x}=-\frac{1}{2}$ (which is a linear equation) 

$\text{ I. }\mathrm{F}.=e^{-2\int \frac{1}{x}dx}=x^{-2}$ 

Multiplying with I.F. and integrating, we have

$\begin{array}{l}y{x}^{-2}=-\frac{1}{2}\int x^{-2}dx+C\\ =\frac{1}{2}{x}^{-1}+C\\ \Rightarrow y=\frac{1}{2}x+C{x}^2\end{array}$