Answer to Question #294882 in Differential Equations for Venkatesh

Question

Solve the equation of x²p²+xp-(y²+y)=0

Answer 1

$\displaystyle x²p²+xp-(y²+y)=0\\\Rightarrow p=\frac{-x\pm\sqrt{x^2-4x^2(-(y^2+y))}}{2x^2}=\frac{-x\pm\sqrt{x^2-4x^2(-y^2-y)}}{2x^2}$

$\displaystyle =\frac{-x\pm\sqrt{x^2+4x^2(y^2+y)}}{2x^2}=\frac{-x\pm\sqrt{x^2(1+4(y^2+y))}}{2x^2}$

$\displaystyle =\frac{-x\pm x\sqrt{(1+4(y^2+y))}}{2x^2}=\frac{-x\pm x\sqrt{4y^2+4y+1}}{2x^2}$

$\displaystyle =\frac{-1\pm \sqrt{4y^2+4y+1}}{2x}$, using quadratic formula.

$\displaystyle \Rightarrow p=\frac{\partial z}{\partial x}=\frac{-1\pm \sqrt{4y^2+4y+1}}{2x}=\frac{-1\pm \sqrt{4y^2+4y+1}}{2x}=\frac{-1\pm\sqrt{(2y+1)^2}}{2x}$

$\displaystyle \Rightarrow \frac{\partial z}{\partial x}=\frac{-1\pm(2y+1)}{2x}= \begin{cases} \frac{y}{x}& \text{if }\frac{\partial z}{\partial x}=\frac{-1+(2y+1)}{2x}\\ \quad\\ \frac{-1-y}{x}& \text{if }\frac{\partial z}{\partial x}=\frac{-1-(2y+1)}{2x} \end{cases}$

$\displaystyle \Rightarrow z= \begin{cases} \int\frac{y}{x}\ dx=y\int\frac{1}{x}\ dx=y\ln x+f(y)\\ \quad\\ \int\frac{-(1+y)}{x}\ dx=-(1+y)\int\frac{1}{x}\ dx=-(1+y)\ln x+ f(y) \end{cases}$

where $f(y)$ is an arbitrary function of $y$.

Thus, $\displaystyle z=y\ln x+f(y) \text{ or }z=-(1+y)\ln x+f(y)$