Question

y=2px-p^2 where p=dy/dx

Answer 1

y = p(2x-p)
y = y/x(2x - y/x)
1 = 1/x (2x^2 -y/x)
x^2 = 2x^2 -y
-x^2 = -y
x^2 = y
x = √y

Answer 2

We need to recall the following concept of the integrating factor method.

When a differential equation is present in the form $\frac{dy}{dx} +P(x)y=Q$ then

Integrating factor is $I.F=e^{\int\(P(x) dx }$

This problem is about finding the solution using an integrating factor.

Given:

$y=2px-p^2$ where $p=\frac{dy}{dx}$

Differentiate the given equation w.r.t. $x$

$\frac{dy}{dx}=2p+2x \frac{dp}{dx}-2p\frac{dp}{dx}$

$p=2p+2x \frac{dp}{dx}-2p\frac{dp}{dx}$

$2x \frac{dp}{dx}-2p\frac{dp}{dx}+p=0$

$2(x-p)\frac{dp}{dx}+p=0$

$(x-p)\frac{dp}{dx}+\frac{p}{2} =0$

$\frac{dx}{dp}+\frac{2}{p}x =2$

Here,  $P=\frac{2}{p}$ and $Q=2$

So, the integrating factor is,

$I.F=e^{\int\(\frac{2}{p} dp }=e^{2logp}=p^2$

Therefore, the solution to the equation is as follows

$x(I.F)=\int\l.F(Q) dp$

$xp^2=\int\(p^2(2) dp$

$xp^2=\frac{2}{3} p^3+C$

⇒ $x=\frac{2}{3}p+Cp^{-2}$