Differential Equation - should be simple, but?

[QuasarRage]

Homework Statement

I am studying for the exams, and not terribly practiced on DE's yet. From a mock-test I found this DE - it be solved? A complete solution would be appreciated so I can understand the answer and compare it to my notes please...


dy/dx - 2xy = ( y ln (y) ) / x

The Attempt at a Solution

I have attempted so far:

a) Separate variables.
I cannot, I get stuck at something like dy + ( -2XY -(Y ln(y)) ) dx = 0, which I tried to make dy*x/y + (-2x^2 -lny )dx = 0 and cannot find a way to separate further

b) Find a multiplicator (is it how is it called? sorry I am not a native English speaker and not yet very familiar with English mathematics terminology), anyway, something to multiply it with so that it can become directly differentable. I don't know how to go about it at the moment.

c) Find a suitable transformation. No luck...

Thanks for reading, thanks for the solution if you provide it...

 

[Dick]

Write it as:

(x/y)*dy-(2*x^2+ln(y))*dx=0 and think about it again. "Multiplicator"="Integrating factor".

 

[QuasarRage]

Thanks for the answer. I actually reached that point, but unless I am a complete dummy, it still is not an exact DE.

...M(x,y)....N(x,y)
(2x^2 + lny)dx + (-x/y)dy = 0

dM/dy = 1/y != dN/dx = -1/y so it is not exact.

.So how do I tackle the minus? Is it still a matter of finding a correct integrating factor? Please take it a step further for me, this is actually the point I am stuck at, it is so frustrating I cannot describe it

 

[Dick]

You are right. I missed the sign. Back to the drawing board.

 

[mybsaccownt]

i hope it's not bad form to re-use my post from a similar type of question:from this point:

...M(x,y)....N(x,y)
(2x^2 + lny)dx + (-x/y)dy = 0

dM/dy = 1/y != dN/dx = -1/y so it is not exact.

.So how do I tackle the minus? Is it still a matter of finding a correct integrating factor? Please take it a step further for me, this is actually the point I am stuck at, it is so frustrating I cannot describe it

yes, you can convert it to exact by trying:

Yes...it does look like it would be an exact equation, but it's not quite there.

Is there anything we can do to make it so?

how about finding an integrating factor to multiply through in order to convert it to exact?

if...

\frac{My-Nx}{N}

is a function of x only, then the solution to:

\frac{d\mu}{dx} = \frac{My-Nx}{N} \mu

gives \mu as the appropriate integrating factor

or if

\frac{Nx-My}{M}

is a function of y only, then the solution to:

\frac{d\mu}{dx} = \frac{Nx-My}{M} \mu

gives you the integrating factor, \mu

 

[dextercioby]

SO you have the equation

\frac{dy}{dx}=2xy+\frac{y}{x}\ln y

Make the substitution
y(x)=e^{p(x)}

The resulting ODE is

\frac{dp}{dx}=\frac{p}{x}+2x

For x\neq 0, we get

\frac{1}{x}\frac{dp}{dx}-\frac{p}{x^2}=2

which is the same as

\frac{d}{dx}\left(\frac{p(x)}{x}\right)=2

The rest is trivial.

 

[QuasarRage]

Thank you very much dick, mybsaccownt and dextercioby for the help, I have succeded in solving it both ways proposed (without looking at your solutions while I was doing it ofc :) )

My solution was: -2x + y + (lny)/x = c

Thanks again very much for your help.

For dextercioby's solution, what hinted us to this transformation pls? Was it the presence of lny in the equation?

 

[dextercioby]

Yes, of course, that way, i got rid of a nasty function of "y".