Solve this DE using homogeneous equations

[BifSlamkovich]

Homework Statement

dy/dx = (6x^(2)+xy+6y^(2))/(x^2)

Homework Equations

v = y/x
y' = v + xv'

The Attempt at a Solution

y' = tan(6ln(abs(x))-C)/x ===> apparently not correct

 

[Curious3141]

Homework Statement

dy/dx = (6x^(2)+xy+6y^(2))/(x^2)

Homework Equations

v = y/x
y' = v + xv'

The Attempt at a Solution

y' = tan(6ln(abs(x))-C)/x ===> apparently not correct

Tough to see where you went wrong if you don't show all your working.

In your final line, I presume you meant $y$ instead of $y'$ (the latter means derivative of $y$).

Your answer looks almost correct, except that in your last step in going from $v$ to $y$ you may have divided by $x$ instead of multiplying by $x$. Remember that $v = \frac{y}{x}$, so $y = vx$.

Another pointer (not that your post is wrong in this respect, but this makes things neater) is that when you have a ln term plus an arbitrary constant, it's neater to express it as $\ln|kx|$. This is because $\ln|kx| = \ln|x| + \ln|k|$, which is equivalent in form to $\ln|x| + C$.

So your final answer should be $y = x\tan(6\ln|kx|)$, where k is an arbitrary constant.

PS: also, to correct a possible misconception in your thread title, you're not solving the DE "using" homogeneous equations. The DE itself is a homogeneous equation, the definition of which is that multiplying all variables (both $x$ and $y$ in this case) by a constant scaling factor will not change the equation.

 

[HallsofIvy]

You are correct up to the very last step. y= xv, not v/x!