Question

Find solution for $DE\Rightarrow {\left(\frac{dy}{dx}\right)}^2-y\frac{dy}{dx}+x=0$.

Answer 1

Given,

${\left(\frac{dy}{dx}\right)}^2-y\frac{dy}{dx}+x=0$

substitute $\frac{dy}{dx}=p$

$p^2-py+x=0$

$\Rightarrow x=py-p^2$.....(1)

using Clairaut's equation, differentiate both sides w.r.t p

$\frac{dx}{dp}=y-2p$

to find solution put $\frac{dx}{dp}=0$

$0=y-2p$

$p=\frac{y}{2}$

substitute value of p in equation (1)

$\Rightarrow x=\left(\frac{y}{2}\right)y-{\left(\frac{y}{2}\right)}^2$

$x=\frac{y^2}{2}-\frac{y^2}{4}$

$\therefore x=\frac{y^2}{4}$

Is the required equation.