Mathematics - Differential Equation Question with Solution | TestHub

Let $P(x,y)$ be any point on the curve. Length of intercept on y -axis by any tangent at P is

$OT=y-x\frac{dy}{dx}\frac{O{O}^{l}P(x,y)}{L}$

∴ Area of trapezium $OLPTO=\frac{1}{2}(PL+OT)OL$

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

According to question

Area of trapezium $OLPTO=\frac{1}{2}{x}^2$

i.e., $\frac{1}{2}\left(2y-x\frac{dy}{dx}\right)x=\pm \frac{1}{2}{x}^2$

$\Rightarrow 2y-x\frac{dy}{dx}=\pm x$ or $\frac{dy}{dx}-\frac{2y}{x}=\pm 1$

Which is linear differential equation and I.F. $=e^{-2\ln x}=\frac{1}{x^2}$

∴ The solution is $\frac{y}{x^2}=\int \pm \frac{1}{x^2}dx+c=\pm \frac{1}{x}+c$

$\therefore y=\pm x+c{x}^2$ or $y=c{x}^2\pm x$, where c is an arbitrary constant