obtain a family of solution for $$3(3x^2 + y^2)dx -2xydy = 0$$ i just need this as basis for how the answer will look like but i still need steps

Question

cwtan · Answer

sub y=vx
and dy/dx=v+x dv/dx?

cwtan · Answer

I've 4gt how to integrate..... lol

lgbasallote · Answer

i think exact DE can be used though

lgbasallote · Answer

ugh it cant =_= isnt there any other way for this >.<

cwtan · Answer

This is what i have done:
$$3(3x^2+y^2)dx=2xy dy$$
$$\frac{dy}{dx}=\frac{3(3x^2+y^2)}{2xy}$$
y=vx
$$v+x\frac{dv}{dx}=\frac{3(3x^2+v^2x^2)}{2vx^2}$$
$$v+x\frac{dv}{dx}=\frac{3(3+v^2)}{2v}$$
$$x\frac{dv}{dx}=\frac{3(3+v^2)-2v^2}{2v}$$
$$x\frac{dv}{dx}=\frac{9-v^2}{2v}$$
$$\frac{1}{x}dx=\frac{2v}{9-v^2}dv$$
ok following is integration which is the part i duno.........

lgbasallote · Answer

$$\ln x = -\ln |9 - v^2| + c$$
$$\ln |x(9 - v^2)| = C$$

binary3i · Answer

x(9-v^2)=c
x9-y^2/x =c

lgbasallote · Answer

$$\ln |x(\frac{9x^2 - y^2}{x^2})| = C$$ $$\ln |\frac{9x^2 - y^2}{x}| = C$$

lgbasallote · Answer

$$\frac{9x^2 - y^2}{x} = C_1$$ seems good to me..

cwtan · Answer

yea u've done it