Solve x^2 (dy/dx)^2-2xy dy/dx+2y^2-x^2=0 | Equation Solvable for p

 Equation Solvable for p 


MathJax Example

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

Solution:

\[ \begin{align*} & \text{The given equation is} \\ & x^2 \left(\frac{dy}{dx}\right)^2 - 2xy \frac{dy}{dx} + 2y^2 - x^2 = 0 \\ & \text{or} \\ & \left(\frac{dy}{dx}\right)^2 - 2xy \frac{dy}{dx} + 2y^2 - x^2 = 0 \\ & \text{or} \\ & \left(\frac{dy}{dx}\right) = \frac{2xy \pm \sqrt{4x^2 y^2 - 4x^2 (2y^2 - x^2)}}{2x^2} \\ & \text{or} \\ & \left(\frac{dy}{dx}\right) = \frac{y \pm \sqrt{x^2 - y^2}}{x} \\ \end{align*} \]

Put \(y = vx\) so that \(\frac{dy}{dx} = v + x \frac{dv}{dx}\).

Therefore, (2) becomes \(v + x \frac{dv}{dx} = \frac{vx \pm \sqrt{x^2 - v^2 x^2}}{x}\).

Therefore, \(v + x \frac{dv}{dx} = v \pm \sqrt{1 - v^2}\).

Therefore, \(x \frac{dv}{dx} = \pm \sqrt{1 - v^2}\).

Separating the variables, \(\frac{1}{\sqrt{1 - v^2}} dv = \pm \frac{1}{x} dx\).

MathJax Example

Integrating, \(\int \frac{1}{\sqrt{1-v^2}} \, dx = \pm \int \frac{1}{x} \, dx\).

Taking the positive sign, we get:

\[ \sin^{-1} v = \log |x| + \log |c_1| \\ \text{or} \\ \sin^{-1} \frac{y}{x} = \log |c_1 x| \\ \therefore \, |c_1 x| = e^{\sin^{-1} \frac{y}{x}} \\ \text{or} \\ \pm c_1 x = e^{\sin^{-1} \frac{y}{x}} \]

Therefore, \(\pm e^{\sin^{-1} \frac{y}{x}} - c_1 x = 0\) where \(\pm c_1 = c\).

Taking the negative sign, we get:

\[ -\sin^{-1} v = \log |x| + \log |c_2| \\ \text{or} \\ \pm c_2 x = e^{-\sin^{-1} \frac{y}{x}} \]

Therefore, the general solution of (1) is \((e^{\sin^{-1} \frac{y}{x}} - c x)(e^{-\sin^{-1} \frac{y}{x}} - c x) = 0\).

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