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

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

pls show all working

To find the general solution of the given first-order differential equation, we can follow these steps:

Step 1: Rearrange the equation
Rewrite the given equation as: du/u = dt/(t + 2t)

Step 2: Simplify the equation
Combine the terms on the right side of the equation: du/u = dt/(3t)
Simplify the right side by combining like terms: du/u = dt/(3t)
Simplify further: du/u = (1/3) * dt/t

Step 3: Integrate both sides
Now, we integrate both sides of the equation.
Integrate the left side with respect to u: ∫(1/u) du = ∫(1/3t) dt

The integral of (1/u) du is ln|u| + C1, where C1 is the constant of integration.
The integral of (1/3t) dt is (1/3) * ln|t| + C2, where C2 is another constant of integration.

Therefore, we have the equation: ln|u| + C1 = (1/3) * ln|t| + C2

Step 4: Combine constants
Combine the constants on the right side of the equation: ln|u| + C1 = ln|t|/3 + C2

Step 5: Convert to exponential form
Rewrite the equation using exponential form:
ln|u| = ln|t|/3 + C2 - C1

Applying the logarithmic property that ln(a) - ln(b) = ln(a/b), we have:
ln|u| = ln|t|/3 + C2 - C1
ln|u| = ln|t|/3 + C3, where C3 = C2 - C1

Step 6: Eliminate logarithms
Remove the natural logarithm on both sides by exponentiating. Exponentiating both sides with base e gives us:
|u| = e^(ln|t|/3 + C3)

Using the property that e^(ln x) = x, we have:
|u| = |t|/3 * e^(C3)

Step 7: Account for absolute value
Consider both the positive and negative cases for the absolute value:
For |u| = |t|/3 * e^(C3), if u > 0, then u = t/3 * e^(C3)
If u < 0, then u = -t/3 * e^(C3)

Therefore, the general solution to the first-order differential equation is:
u(t) = C * t/3 * e^(C3), where C is an arbitrary constant, and t ≠ 0.