find the general solution of the following first order differential equation.

u(t): du/dt = u /t+2t

pls show all working

I assume you mean

du/dt = (u/t) + 2t and not
u/(t + 2t) = u/(3t)

Your D.E. is of the form
du/dt + Pu = Q
where P = -1/t and Q = 2t

The method of "integrating factors" can be used. Your textbook should have an explanation of the method.

There is a function rho(t) given by
rho(t) = exp(integral of P(t)dt)

such that the solution is
rho(t)*u
= integral of (rho*Q) + C

To find the general solution of a first-order differential equation, in this case, du/dt = u/t + 2t, we will follow these steps:

1. We will start by separating the variables. Move the u term to one side of the equation and the t term to the other side:

du/u = (1/t)dt + 2tdt

2. Next, we will integrate both sides of the equation. On the left side, the integral of (du/u) is ln|u|. On the right side, we will integrate each term separately:

∫(1/t)dt + ∫2tdt = ln|u|

ln|t| + t^2 + C = ln|u|

(where C is the constant of integration)

3. Now, we will exponentiate both sides of the equation (using the property of logarithms):

e^(ln|t| + t^2 + C) = e^ln|u|

t * e^(t^2) * e^C = |u|

e^C is an arbitrary constant, so let's combine it with a new constant D:

D = e^C

t * e^(t^2) * D = |u|

4. Finally, we can write the general solution by considering the absolute value:

u(t) = ±t * e^(t^2) * D

That is the general solution to the given first-order differential equation, where D is an arbitrary constant that can be determined by initial conditions or any additional constraints on the problem.