Answer to Question #288874 in Differential Equations for Pankaj

Question

Solve :

P² -2xyp + 4y²=0, where p=dy/dx

Answer 1

Using quadratic formula;

$\displaystyle p=\frac{-(-2xy)\pm\sqrt{(-2xy)^2-4(1)(4y^2)}}{2(1)}\\ \quad=\frac{2xy\pm\sqrt{4x^2y^2-16y^2}}{2}\\ \quad=\frac{2xy\pm\sqrt{4y^2(x^2-4)}}{2}\\ \quad=\frac{2xy\pm2y\sqrt{x^2-4}}{2}\\ \quad=xy\pm y\sqrt{x^2-4}=y(x\pm\sqrt{x^2-4})\\$

But, $\displaystyle p=\frac{dy}{dx}\\$, thus we have;

$\displaystyle \frac{dy}{dx}=y(x\pm\sqrt{x^2-4})\\ \Rightarrow\frac{1}{y}\frac{dy}{dx}=x\pm\sqrt{x^2-4}$ , by method of separation of variable

Integrating both sides wrt x yields;

$\displaystyle \int\frac{1}{y}\frac{dy}{dx}dx=\int(x\pm\sqrt{x^2-4})dx\\ \int\frac{1}{y}dy=\int x\ dx\pm\int\sqrt{x^2-4}\ dx\qquad\cdots\cdots\cdots\cdots\cdots\cdots\cdots(1)$

But,

$\displaystyle \int\sqrt{x^2-4}\ dx=4\int\sinh^2p\ dp$ , where x=2coshp

$\displaystyle =2\int(\cosh 2p-1)\ dp$ , since $\displaystyle \cosh 2p=1+2\sinh^2p$

$\displaystyle =\sinh 2p-2p=2\sinh p \cosh p=2\times\frac{\sqrt{4-x^2}}{2}\times\frac{x}{2}-2 \cosh^{-1}\left(\frac{x}{2}\right)$

$\displaystyle =\frac{x\sqrt{x^2-4}}{2}-2\cosh^{-1}\left(\frac{x}{2}\right)\qquad\cdots\cdots\cdots\cdots(2)$

Substituting (2) into (1) yields;

$\displaystyle \log y=\frac{x^2}{2}\pm\left[\frac{x\sqrt{x^2-4}}{2}-2\cosh^{-1}\left(\frac{x}{2}\right)\right]+C, \text{where C is an arbitrary constant}\\ \Rightarrow y=Ae^{\left(\frac{x^2}{2}\pm\left[\frac{x\sqrt{x^2-4}}{2}-2\cosh^{-1}\left(\frac{x}{2}\right)\right]\right)}, \text{where }A=e^C$