Solving Differential Equation: {x^2} - 3{y^2}+6xy

[drawar]

Sorry I couldn't think of any more relevant title. Here's the equation:

${x^2} - 3{y^2} + 6xy\frac{{dy}}{{dx}} = 0$

I'm thinking of rewriting the above to $\frac{{dy}}{{dx}} = \frac{{3{y^2} - {x^2}}}{{6xy}}$ followed by a change of variable u=y/x. But should I rule out the case when either x=0 or y=0 first? I'd also love to see if there's any alternative way to solve this ODE, thanks!

 

[D H]

That term containing y*dy/dx suggests you should consider the substitution u=y².

 

[Chestermiller]

That term containing y*dy/dx suggests you should consider the substitution u=y².

Yes! And also let v = x². Then see how these substitutions simplify things.

Chet

 

[HallsofIvy]

Sorry I couldn't think of any more relevant title. Here's the equation:

${x^2} - 3{y^2} + 6xy\frac{{dy}}{{dx}} = 0$

I'm thinking of rewriting the above to $\frac{{dy}}{{dx}} = \frac{{3{y^2} - {x^2}}}{{6xy}}$ followed by a change of variable u=y/x. But should I rule out the case when either x=0 or y=0 first? I'd also love to see if there's any alternative way to solve this ODE, thanks!

Notice that
$$\frac{3y^2- x^2}{6xy}= \frac{\frac{3y^2- x^2}{xy}}{6}= \frac{1}{6}(3\frac{y}{x}- \frac{x}{y})$$
Which suggests that the substitution v= y/x would be useful (actually, the fact that both numerator and denominator are of second degree first suggested that).

If v= y/x then y= xv so that y'= xv'+ v. The differential equation becomes
$$x\frac{dv}{dx}+ v= \frac{1}{6}\left(3v- \frac{1}{v}\right)$$
$$x\frac{dv}{dx}= \frac{1}{6}\left(3v- \frac{1}{v}\right)- v= \frac{1}{6}\left(-3v- \frac{1}{v}\right)$$
$$x\frac{dv}{dx}= -\frac{1}{6}\left(\frac{3v^2+ 1}{v}\right)$$
$$\frac{v}{3v^2+ 1}dv= -\frac{1}{6x}dx$$