How to solve this ordinary differential equation?
$$ (3x+5)\Big(\frac{\mathrm{d}y}{\mathrm{d}x}\Big)^2-(3y+x)\Big(\frac{\mathrm{d}y}{\mathrm{d}x}\Big)+y=0 $$
How to find the general solution and singular solution of this differential equation?
Thank you in advance for your help!
Let $p = \frac{dy}{dx}$. Therefore the differential equation is, $$(3x+5)p^2-(3y+x)p+y=0$$ Differentiating with respect to x, $$3p^2+6xp\frac{dp}{dx}+10p\frac{dp}{dx}-3p^2-3y\frac{dp}{dx}-p-x\frac{dp}{dx}+p=0$$ $$\therefore 6xp\frac{dp}{dx}+10p\frac{dp}{dx}-3y\frac{dp}{dx}-x\frac{dp}{dx}=0$$ $$\therefore \frac{dp}{dx}=0 \implies p =C$$ where C is an arbitrary constant or, $$6xp+10p-3y-x=0 \implies p=\frac{3y+x}{6x+10}$$ Substituting the first relation in the original equation gives us the family of curves, $$y=\frac{(3x+5)C^2-Cx}{3C-1}$$ And substituting the second relation gives us the singular solution to be $$y=\frac{(x+3y)^2}{4(3x+5)}$$