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
$$x\frac{dv}{dx}+ v= \frac{1}{6}\left(3v- \frac{1}{v}\right)$$
$$x\frac{dv}{dx}= \frac{1}{6}\left(3v- \frac{1}{v}\right)- v= \frac{1}{6}\left(-3v- \frac{1}{v}\right)$$
$$x\frac{dv}{dx}= -\frac{1}{6}\left(\frac{3v^2+ 1}{v}\right)$$
$$\frac{v}{3v^2+ 1}dv= -\frac{1}{6x}dx$$

 

[blue_raver22]

There are actually multiple ways to approach this differential equation. One method is to use the substitution u=y/x, as you mentioned. This will lead to the equation du/dx = (3u^2 - 1)/(6u). From here, you can separate variables and integrate to find the general solution. However, as you pointed out, this method will not work if either x=0 or y=0. To account for this, you can split the equation into two cases: x=0 and y=0, and solve each separately.

Another approach is to use the method of exact differential equations. This involves finding a function u(x,y) such that du/dx = x^2 - 3y^2 + 6xy. This can be done by integrating with respect to x and then setting the resulting expression equal to u. Once u is found, the solution can be found by setting u=C, where C is a constant.

In terms of ruling out the case of x=0 or y=0, it is important to consider the domain of the original equation. Depending on the context of the problem, these cases may not be relevant and can be omitted. However, if x=0 and y=0 are potential solutions that need to be considered, then they should be included in the solution process.

Overall, there are various ways to approach and solve this differential equation. It is important to consider the domain of the equation and use appropriate methods to find the solution.