Integral respect to time What is the integral of with respect to time $\frac{d[c]}{dt} = k[c]$?
It's said that it will be $c = k[c]t + c_0$
but because we are making integral based on $t$ why should I add initial constant $c_0$ here?
 A: The solution only makes sense iff
$$
\dfrac{dc}{dt} = k_c
$$
where $k_c$ is a constant
leading to
$$
c(t) = k_c t + C
$$
setting t = 0 we find
$$
c(t=0) = c(0) = k_c \cdot 0 + C =\implies c(0) = C
$$
thus
$$
c(t) = k_c t + c(0)
$$
if I am assuming too much then correct me where I am wrong.
A: In general the antiderivative of a function is determined only up to a constant, and this constant can be fixed by specifying the value of the antiderivative you want at a fixed time.
In your case, your differential equation is equivalent to
$$\frac{d[c]}{[c]} = k\, dt.$$
Integrating both sides, and consolidating the constants on the RHS gives
$$\log[c] = kt + C.$$
Since concentration $[c]$ is presumably a positive quantity, solving for $[c]$ gives
$$[c](t) = e^{kt + C} = C' e^{kt},$$
where $C' := e^C$. We can solve for $C'$ by, for example, evaluating at $t = 0$:
$$[c](0) = C'.$$
Substituting gives
$$[c](t) = [c](0) e^{kt}.$$
A: Sounds like you've just started A-level chemistry, and because you've done AS maths you're trying to solve the rate equations you've just been taught about formally. I'm assuming this is true so I'm going to solve $\frac{d[c]}{dt}=k[c]$ for you in this context.
Firstly, we write $\frac{1}{[c]}*\frac{d[c]}{dt}=k$. Now, you probably haven't seen it before but $\frac{d(ln x)}{dt}  = \frac{1}{x}\frac{dx}{dt}$. Using this result, we can write $\frac{d(ln[c])}{dt}=k$ and we can integrate that (just write $ln[c]=y$ if it looks too weird) to give $ln[c]=kt+A$. Now, our initial conditions are $[c]$ at $t=0$ is $c_0$. so $A=ln(c_0)$ i.e. $ln[c]=kt+ln(c_0)$.
To make a connection with the concentration time graph, we need to re-arrange this to $[c]=f(t)$. You've probably seen logs to base 10, $ln$ is logs to base $e$.
$\exp(ln[c])=\exp(kt+ln(c_0))$
$[c]=\exp(kt)\exp(ln(c_0))$
$[c]=c_0 \exp(kt)$
Note that this relies on $c_0$ not being equal to $0$ - $ln(0)$ is undefined. This equation is typically useful for reactions where $[c]$ is a reactant, not a product. In this case, k will be negative. This then shows exponential decay from $c_0$ to, eventually, nothing.
To get the equation for the growth of the product (which i will call $P$), we need to look at the chemical equation. If we have C -> P we determine that $\frac{d[c]}{dt}=-\frac{d[P]}{dt} and then solve.
$\frac{d[P]}{dt}=-k[c]=-kc_0\exp(kt)$
Using another standard result
$[P]= -c_0\exp(kt) + B$
the initial condition this time are $[p]=0$ at $t=0$ so
$[P]=c_0(1-\exp(kt))$
This shows growth from nothing to, eventually, c_0, which is consistent with the chemical equation I wrote down. If instead we had C->2P, then we would have used the rate equation $2\frac{d[c]}{dt} = -\frac{d[P]}{dt}$ i.e. $[P]$ grows at twice the rate of $[c]$ decreasing. The solution should then be $[P]=2c_0(1-\exp(kt))$ to be self consistent.
A: Here are the steps
$$ \frac{d}{dt}[c]=k[c] $$
$$ \frac{1}{[c]}d[c]=k\ dt $$
$$ \int \frac{1}{[c]}d[c]=k\int dt $$
$$ \ln |[c]|+a_1=kt+a_2 $$
$$ \ln |[c]|=kt+a_2-a_1=kt+a $$
$$ e^{\ln |[c]|}=e^{kt+a} $$
$$ [c]= e^{kt+a} $$
