How to solve the differential equation $\frac {dP}{dt} = \frac {1}{2} (P-200)$ when we are given that $p=200$ when $t=0$? The title pretty much describes the question. Posting this on behalf of a friend. He says he knows how to apply the integrating factors method, but can't figure out how to solve this.
Here's my working:
$$\frac {dP}{dt} = \frac {1}{2} (P-200)$$
$$ \int \frac {1}{P-200} dP = \int \frac {1}{2} dt $$
$$ ln(P-200) = \frac {1}{2} t + c $$ where $c$ is the constant of integration.
Now, if we substitute the values of $p$ and $t$ in the equation, we would get $$c = ln(0)$$
This is where I'm stuck. This is a JEE past paper question.
I would really appreciate it if someone could tell me where my friend and I are going wrong. Thanks!
 A: Notice by inspection that the constant function $P(t) = 200$ satisfies the IVP. Indeed, it's an unstable equilibrium solution. There is an existence-uniqueness theorem that guarantees that this is the desired solution.
You make a mistake when you divide both sides by $(P - 200)$, since you are dividing by zero.
A: You have divided by $P-200$, which assumes that $P \neq 200$.  By inspection, $P=200$ solves the equation.
A: We have $P'=\frac12 P-100$ so $P_h(t)=Ce^{\frac t2}$ is the homogeneous solution and $200$ is a particular solution so the general solution is 
$$P(t)=Ce^{\frac t2}+200$$
and with $P(0)=200$ we get $C=0$.
A: $$\frac{\text{d}\text{P}(t)}{\text{d}t}=\frac{1}{2}\left(\text{P}(t)-200\right)\Longleftrightarrow$$
$$\text{P}'(t)=\frac{\text{P}(t)-200}{2}\Longleftrightarrow$$
$$\frac{\text{P}'(t)}{\frac{\text{P}(t)-200}{2}}=1\Longleftrightarrow$$
$$\int\frac{2\text{P}'(t)}{\text{P}(t)-200}\space\text{d}t=\int1\space\text{d}t\Longleftrightarrow$$
$$2\ln\left|\text{P}(t)-200\right|=t+\text{C}\Longleftrightarrow$$
$$\text{P}(t)=200+e^{\frac{\text{C}+t}{2}}\Longleftrightarrow$$
$$\text{P}(t)=200+\text{C}e^{\frac{t}{2}}$$
Now, knowing that $\text{P}(0)=200$:
$$200=200+\text{C}e^{\frac{0}{2}}\Longleftrightarrow$$
$$200=200+\text{C}e^{0}\Longleftrightarrow$$
$$200=200+\text{C}\cdot1\Longleftrightarrow$$
$$200=200+\text{C}\Longleftrightarrow$$
$$200-200=\text{C}\Longleftrightarrow$$
$$\text{C}=0$$
So, the solution is:
$$\text{P}(t)=200$$
