Answer to Question #130626 in Differential Equations for Jijo

Question

x^2p^²-2xyp+2y^2-x^2=0
Using Claratiurs method

Answer 1

The equation above is not reducible to the Clairaut's Form as it can't be put in the form

$y = xp +f(p)$

To solve this equation, simple use $x^2p^2-2xyp+y^2-x^2 = 0$

Factorizing the above equation.

we get, $y = xp - x$ and $y = xp +x$

Solving the equations,

$y = x(\frac{dy}{dx} - 1) \implies \frac{dy}{dx} - \frac{y}{x} = 1$

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

Then,

$\frac{1}{x} y = \int \frac{1}{x} dx$

$\frac{1}{x} y = lnx + ln C$

$y = x(lnx + ln C) = xln(Cx)$

Similarly for $y = x(p+1) \implies \frac{dy}{dx} - \frac{y}{x} = -1$

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

$\frac{1}{x} y = \int -\frac{1}{x} dx$

$y = -x(lnx + ln C) = -xln(Cx)$