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

Find a one parameter family of solutions of the following first order ordinary differential equation
$\left(3{x}^{2}+9xy+5{y}^{2}\right)dx-\left(6{x}^{2}+4xy\right)dy=0$
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plodno8n
The equation is
$3{x}^{2}+9xy\left(x\right)+5y\left(x{\right)}^{2}-\left(6{x}^{2}+4xy\left(x\right)\right){y}^{\prime }\left(x\right)=0$
Looking at the last term, let $y\left(x\right)=u\left(x\right)-\frac{3}{2}x$ to get
$-4xu\left(x\right){u}^{\prime }\left(x\right)+5u\left(x{\right)}^{2}+\frac{3{x}^{2}}{4}=0$
that is to say
$-2x{\left({u}^{2}\left(x\right)\right)}^{\prime }+5{u}^{2}\left(x\right)+\frac{3{x}^{2}}{4}=0$
So, let $u\left(x\right)=±\sqrt{v\left(x\right)}$ to get
$-2x{v}^{\prime }\left(x\right)+5v\left(x\right)+\frac{3{x}^{2}}{4}=0$
which looks to be simple.

telegrafyx
This is a homogeneous DE so making $y=ux$ we obtain
$x{u}^{\prime }=\frac{u\left(u+3\right)+3}{4u+6}$
which is separable giving
$\frac{\left(4u+6\right)du}{u\left(u+3\right)+3}=\frac{dx}{x}$
or
$\mathrm{ln}\left(u\left(u+3\right)+3{\right)}^{2}={C}_{0}+\mathrm{ln}x⇒\left(u\left(u+3\right)+3{\right)}^{2}={C}_{1}x$
and finally
$y=\frac{x}{2}\left(-3±\sqrt{{C}_{2}\sqrt{x}-3}\right)$