Find a one parameter family of solutions of the following first order ordinary differential equation

Question

Find a one parameter family of solutions of the following first order ordinary differential equation

$$(3x^2 + 9xy + 5y^2) dx - (6x^2 + 4xy) dy = 0$$

Hello. So I am stuck after I find out that they are not exact. Please help.

Answer 1

Fist of all, welcome to the site !

The equation is $$3 x^2+9 x y(x)+5 y(x)^2-\left(6 x^2+4 x y(x)\right) y'(x)=0$$ Looking at the last term, let $y(x)=u(x)-\frac 32 x$ to get $$-4 x u(x) u'(x)+5 u(x)^2+\frac{3 x^2}{4}=0$$ that is to say $$-2x \left(u^2(x)\right)'+5 u^2(x)+\frac{3 x^2}{4}=0$$ So, let $u(x)=\pm \sqrt{v(x)}$ to get $$-2 x v'(x)+5 v(x)+\frac{3 x^2}{4}=0$$ which looks to be simple.

Answer 2

Integrating factor is $$\mu=\frac{1}{xP+yQ}\\= \frac{1}{x\, \left( 5 {{y}^{2}}+9 x y+3 {{x}^{2}}\right) +y\, \left( -4 x y-6 {{x}^{2}}\right) }\\= \frac{1}{x\, {{y}^{2}}+3 {{x}^{2}} y+3 {{x}^{3}}}\\ $$ General solution is $$5 \log{(x)}-2 \log{\left( {{y}^{2}}+3 x y+3 {{x}^{2}}\right) }=C$$

Answer 3

This is a homogeneous DE so making $y = u x$ we obtain

$$ x u'=\frac{u(u+3)+3}{4u+6} $$

which is separable giving

$$ \frac{(4u+6)du}{u(u+3)+3} = \frac{dx}{x} $$

or

$$ \ln(u(u+3)+3)^2 = C_0 + \ln x\Rightarrow (u(u+3)+3)^2=C_1x $$

and finally

$$ y = \frac x2\left(-3\pm\sqrt{C_2\sqrt x-3}\right) $$