Hello Trinket,
We are told to model a person's weight with the IVP:
$$\frac{dw}{dt}=\frac{C}{3500}-\frac{17.5}{3500}w$$ where $$w(0)=w_0$$
I would choose the write the ODE in standard linear form:
$$\frac{dw}{dt}+\frac{1}{200}w=\frac{C}{3500}$$
Next, we may compute the integrating factor:
$$\mu(t)=e^{\frac{1}{200}\int\,dt}=e^{\frac{1}{200}t}$$
Multiplying the ODE by the integrating factor, we obtain:
$$e^{\frac{1}{200}t}\frac{dw}{dt}+\frac{1}{200}e^{ \frac{1}{200}t}w=\frac{C}{3500}e^{ \frac{1}{200}t}$$
Now, this allows us the express the left side of the ODE as the differentiation of a product:
$$\frac{d}{dt}\left(e^{\frac{1}{200}t}w \right)=\frac{C}{3500}e^{\frac{1}{200}t}$$
Integrate with respect to $t$:
$$\int \frac{d}{dt}\left(e^{\frac{1}{200}t}w \right)\,dt=\frac{C}{3500}\int e^{\frac{1}{200}t}\,dt$$
$$e^{\frac{1}{200}t}w=\frac{2C}{35}e^{\frac{1}{200}t}+c_1$$
Solve for $w(t)$:
$$w(t)=\frac{2C}{35}+c_1e^{-\frac{1}{200}t}$$
We may determine the parameter $c_1$ by using the initial value:
$$w(0)=\frac{2C}{35}+c_1=w_0\,\therefore\,c_1=\frac{35w_0-2C}{35}$$
And thus, the solution to the IVP is:
(1) $$w(t)=\frac{2C}{35}+\frac{35w_0-2C}{35}e^{-\frac{1}{200}t}$$
The amount of weight lost $L$ is the initial weight minus the current weight, and so we may write:
$$L(t)=w_0-\frac{2C}{35}-\frac{35w_0-2C}{35}e^{-\frac{1}{200}t}$$
$$L(t)=\frac{35w_0-2C}{35}\left(1-e^{-\frac{1}{200}t} \right)$$
Solving this for $t$, we find:
$$\frac{35L(t)}{35w_0-2C}=1-e^{-\frac{1}{200}t}$$
$$e^{-\frac{1}{200}t}=\frac{35\left(w_0-L(t) \right)-2C}{35w_0-2C}$$
(2) $$t=200\ln\left(\frac{35w_0-2C}{35\left(w_0-L(t) \right)-2C} \right)$$
The limiting weight $w_L$ of the person is:
(3) $$w_L=\lim_{t\to\infty}\left(\frac{2C}{35}+\frac{35w_0-2C}{35}e^{-\frac{1}{200}t} \right)=\frac{2C}{35}$$
Now we have formulas to answer the questions.
(A) Find the general solution of the differential equation.
$$w(t)=\frac{2C}{35}+\frac{35w_0-2C}{35}e^{-\frac{1}{200}t}$$
(B) Consider a person who weighs 180lb and begins a diet of 2500 calories per day. How long will it take the person to lose 10lb? 35?
Using the given data:
$$w_0=180,\,C=2500$$
The formula in (2) becomes:
$$t=200\ln\left(\frac{35\cdot180-2\cdot2500}{35\left(180-L(t) \right)-2\cdot2500} \right)=200\ln\left(\frac{1300}{1300-35L(t)} \right)$$
Now to compute the time to lose 10 lbs, we use $L(t)=10$ to get:
$$t=200\ln\left(\frac{1300}{1300-35\cdot10} \right)=200\ln\left(\frac{1300}{950} \right)=200\ln\left(\frac{26}{19} \right)\approx62.7315117710083$$
And to compute the time to lose 35 lbs, we use $L(t)=35$ to get:
$$t=200\ln\left(\frac{1300}{1300-35\cdot35} \right)=200\ln\left(\frac{1300}{75} \right)=200\ln\left(\frac{52}{3} \right)\approx570.526285982664$$
(C) What is the limiting weight of the person?
$$w_L=\frac{2\cdot2500}{35}=\frac{1000}{7}=142. \overline{857142}$$
(D) Repeat (B) for a person who weighs 200lb when the diet started.
Using the given data:
$$w_0=200,\,C=2500$$
The formula in (2) becomes:
$$t=200\ln\left(\frac{35\cdot200-2\cdot2500}{35\left(200-L(t) \right)-2\cdot2500} \right)=200\ln\left(\frac{2000}{2000-35L(t)} \right)$$
Now to compute the time to lose 10 lbs, we use $L(t)=10$ to get:
$$t=200\ln\left(\frac{2000}{2000-35\cdot10} \right)=200\ln\left(\frac{2000}{1650} \right)=200\ln\left(\frac{40}{33} \right)\approx38.4743785294912$$
And to compute the time to lose 35 lbs, we use $L(t)=35$ to get:
$$t=200\ln\left(\frac{2000}{2000-35\cdot35} \right)=200\ln\left(\frac{2000}{775} \right)=200\ln\left(\frac{80}{31} \right)\approx189.607886037747$$
