Question

Find the particular solution W(t) to the differential equation (dW)/(dt)=(1)/(25)(W-300) with the initial condition W(0)=1400.

Answer 1

The particular solution to the differential equation dW/dt = (1/25)(W - 300) with the initial condition W(0) = 1400 is W(t) = 300 + 1100e^(t/25).

Step-by-step explanation:

To solve the differential equation dW/dt = (1/25)(W - 300) with the initial condition W(0) = 1400, we first separate the variables:

1. Rewrite the equation as dW/(W-300) = (1/25)dt.
2. Integrate both sides to get ∗dW/(W-300) = ∗(1/25)dt, which yields ln|W - 300| = t/25 + C.
3. Apply the initial condition to find the constant C. When t=0, W=1400, so ln|1400 - 300| = 0/25 + C, whereby C = ln(1100).
4. Exponentiate both sides to solve for W(t): W(t) = 300 + e^(t/25 + ln(1100)), which simplifies to W(t) = 300 + 1100e^(t/25).

This is the particular solution for W(t).