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.