Question

P^3x+p^2y-p^2x-py=0, where p=dy/dx

Answer 1

Given equation is: $p^3x+p^2y-p^2x-py=0$

$p^2x(p-1) + py(p-1) = 0$

$p(p-1)(px+y) = 0$

Solving it we get, $p=0,p-1 = 0, px+y=0$

Then, for $p = 0, \implies \frac{dy}{dx} = 0$

Integrating it we get, $y = c_1$ (1)

where c₁ is constant.

For p-1=0,

$\frac{dy}{dx} -1= 0$

$\int {dy} =\int dx$

$y = x + c_2$ (2)

For px+y = 0,

$x\frac{dy}{dx} + y= 0$

$x\frac{dy}{dx} = -y$

$\frac{dy}{y} = -\frac{dx}{x}$

integrating both sides,

$\ln y = -\ln x + \ln c_3$

$y = \frac{c_3}{x}$ (3)

Equations (1), (2), and (3) are the solutions of the differential equation.