A hard differential equation - where is the error in my logic?

[Nikitin]

[SOLVED] A hard differential equation - where is the error in my logic?

Homework Statement

2xyy' + y², where y is a function of x

The Attempt at a Solution

2xyy' + y² = 12x² |*(1/x)
=> (y²)' + y²/x =12x |*ln(x), 2yy' = (y²)'
=> (y²ln(x))' = 12x*ln(x)
=> y²ln(x) = ∫12x*ln(x)
=> y²ln(x) = 6x²ln(x) -3x² + C
=> y = ±sqrt[3x²(2-1/ln(x)) + c/ln(x)]

Well, my result is wrong. Can somebody tell me the error in my logic? HELP, my math exam is on the 4th of June...

 

[DryRun]

2xyy' + y² = 12x² |*(1/x)

What is that on the R.H.S. with the vertical bar, etc? Your problem isn't clear.Assuming that your original ODE is: $2xyy'+y^2=12x^2$
Re-arranging gives: y'+\frac{1}{2x}y=6xy^{-1}. You get a Bernoulli equation.

 

[tt2348]

Homework Statement

2xyy' + y², where y is a function of x

The Attempt at a Solution

2xyy' + y² = 12x² |*(1/x)
=> (y²)' + y²/x =12x |*ln(x), 2yy' = (y²)'

=> (y²ln(x))' = 12x*ln(x)
=> y²ln(x) = ∫12x*ln(x)
=> y²ln(x) = 6x²ln(x) -3x² + C
=> y = ±sqrt[3x²(2-1/ln(x)) + c/ln(x)]

Well, my result is wrong. Can somebody tell me the error in my logic? HELP, my math exam is on the 4th of June...

=> (y²)' + y²/x =12x |*ln(x), 2yy' = (y²)'

=> (y²ln(x))' = 12x*ln(x)
(y^2)'*ln(X)+(y^2/x)*ln(x)=/=(y^2*ln(x))'
I believe

 

[Nikitin]

Oh crap, I'm such an idiot! thanks, TT. In addition, I found out how to solve this, so forget about this thread guys!

sharks: I have no idea what a bernoulli equation is, sorry. I haven't even started uni yet.

 

[Ray Vickson]

Homework Statement

2xyy' + y², where y is a function of x

The Attempt at a Solution

2xyy' + y² = 12x² |*(1/x)
=> (y²)' + y²/x =12x |*ln(x), 2yy' = (y²)'
=> (y²ln(x))' = 12x*ln(x)
=> y²ln(x) = ∫12x*ln(x)
=> y²ln(x) = 6x²ln(x) -3x² + C
=> y = ±sqrt[3x²(2-1/ln(x)) + c/ln(x)]

Well, my result is wrong. Can somebody tell me the error in my logic? HELP, my math exam is on the 4th of June...

\frac{d}{dx}(x y^2) = 2 x y y' + y^2.

RGV

 

[dimension10]

I have no idea what a bernoulli equation is

Then just find out!