how do i solve this differentiable equation?

Question

dy/dx=x-2/x



i got an answer of y=x-2lnx+ c and the back of the book says it's y=x-lnx^2 + c and idk how they got the squared. thanks

Answer 1

y = ½ x² − ln (x²) + C

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ln (x²) = 2 ln (x)
This is a basic property of logarithms... from ALGEBRA

Both you and the book are wrong... you found the anti-derivative of 2/x just fine... but not the x.

Answer 2

dT/dt = -ok(T-40) says that the fee of exchange of the temperature is proportional to its distinction from 40 deg. additionally, ok is helpful, so the temperature is changing to flow in the direction of 40 deg. i.e. the better above 40 deg the temperature is, the speedier this is falling, e.g. while the temperature is a hundred deg (i.e. 60 above) this is falling two times as quickly as while this is 70 deg (30 above), and thrice as while it rather is 60 deg (20 above) combine dT/(T - 40) = -ok dt ln(T - 40) = -kt + const. while t = 0, T = a hundred, as a result const = ln 60 for this reason ln(T-40) - ln 60 = -kt ln((T-40)/60) = -kt (T-40)/60 = e^(-kt) while t = 3. T = ninety as a result e^(-3k) = 5/6, ok = (-a million/3) ln(5/6) replace T = 60 into ln((T-40)/60) = -kt: -kt = ln(20/60) = ln(a million/3) as a result t = [ln(a million/3)] / [(a million/3)ln(5/6)] . = 18.077 So it cools to 60 deg in 18 mins (or approximately 1085 s to the closest total style of seconds)

Answer 3

Next time add parentheses

dy/dx=(x-2)/x

y= x - 2ln(x) +C
note that 2ln(x) = ln(x^2)

y = x -ln(x^2) + C

Answer 4

dy/dx=x-2/x
y=∫(x-2/x)dx
=x^2/2-2ln(x)+c
If you differentiate the above, you get x-2/x

If you differentiate what your book says, you get 1-2ln(x)/x

Answer 5

just copy the back of the book then hehe

Answer 6

ln(x^2) = 2 ln(x)