I have the following equation: $$ \frac{dR}{dt}=k_{o}(1-R)-k_{a}R $$
and I want to integrate the equation, whose solution is:
$$ R|_{t}=\frac{k_{o}}{k_{o}+k_{a}}+e^{-(k_{o}+k_{a})t}(R|_{0}-\frac{k_{o}}{k_{o}+k_{a}}) $$
Now, I can get to
$$ R|_{t}=k_{0}(t-\frac{R^{2}}{2})-k_{a}\frac{R^{2}}{2}-C $$
which I hope is correct, but I dont know how to proceed from here mostly to get to the exponentiation part (but I know how to compute C). Can you help me out please?
On
You need to separate the variables R and t.
From the initial equation
$\frac{dR}{dt}= k_o(1−R)−k_a R$
Get all the R's on one side with the $dR$ differential and anything related to $t$ on the other side. Doing this you get:
$\frac{dR}{k_o(1−R)−k_a R} = dt$
Now you can integrate this
$\int \frac{dR}{k_o(1−R)−k_a R} = \int dt$
This can be simplified a bit by collecting the R terms and the constants in the denominator
$\int \frac{dR}{k_o-(k_o+k_a)R} = \int dt$
This is in the form:
$\int \frac{1}{a+bR} dR = \int dt$
The solution of your linear first order differential equation will proceed as follows:
$$\frac{dR}{dt}=k_{o}(1-R)-k_{a}R$$ $$\frac{dR}{dt}=k_{o}-(k_{o}+k_{a})R$$ $$\frac{dR}{dt}+(k_{o}+k_{a})R=k_{o}$$ The integrating factor in this case is $e^{\int (k_{o}+k_{a})dt }=e^{(k_{o}+k_{a})t}$
So multiplying by the integrating factor, we get
$$e^{(k_{o}+k_{a})t}\frac{dR}{dt}+(k_{o}+k_{a})Re^{(k_{o}+k_{a})t}=k_{o}e^{(k_{o}+k_{a})t}$$ $$\frac{d}{dt}(Re^{(k_{o}+k_{a})t})=k_{o}e^{(k_{o}+k_{a})t}$$ $$d(Re^{(k_{o}+k_{a})t})=k_{o}e^{(k_{o}+k_{a})t}dt$$ Integrating both sides w.r.t $t$, we have $$Re^{(k_{o}+k_{a})t}=\frac{k_{o}}{k_{o}+k_{a}}e^{(k_{o}+k_{a})t}+C$$
Thus the general solution is $$R(t)=\frac{k_{o}}{k_{o}+k_{a}}+Ce^{-(k_{o}+k_{a})t}$$
The differential equation you have is of first order and hence your solution contains one constant of integration.
To uniquely determine the constant, you need a constraint known as a boundary condition. According to the condition given, it will become an initial value problem or a boundary value problem.