How do I solve this differential equation

du/dx= 3x- (u/x)

it's a non- separable equation but I have to turn it into a separable one and solve. my lecturer did it by letting u(x)= x^2 + y(x) and building a new equation but I have no idea how this works!

please help!

I write relation as u=x^2+y, then du/dx=2x+dy/dx
so 3x-(u/x)=2x+dy/dx,
3x-(x + y/x)=2x+dy/dx giving
-(y/x)=dy/dx and variables can now be separated.

This answer has made this too complicated, it is much simpler than that.

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Notice that this equation is of the form: du/dx + p(x)u = q(x). The general solution is:

u = ce^(-∫pdx) + e^(-∫pdx) ∫qe^(∫pdx) dx

du/dx + (1/x)u = 3x, where p = (1/x) and q = 3x.

If you work through the quadratures, this should give you the solution. Substitutions like your instructor indicated are not necessary.

There is no need for brackets around the second term, division comes first by default.

Solve this differential equation by using an integrating factor:
du / dx = 3x - u / x
du / dx + u / x = 3x
du / dx + P(x)u = f(x)
P(x) = 1 / x
f(x) = 3x
I(x) = ℮^[∫ P(x) dx]
I(x) = ℮^(∫ 1 / x dx)
I(x) = ℮^lnx
I(x) = x
I(x)u = ∫ I(x)f(x) dx
xu = ∫ 3x² dx
xu = x³ + C
u = x² + C / x

dy/dx=3x-(y/x), put y/x=v, v+xdv/dx=dy/dx

v+xdv/dx=3x-v

xdv/dx=3x-2v