Find y in terms of $t$ So I am given this equation for the rate of a reaction:
$$\frac{dy}{dt}=y^2-8y+15$$ when $t=0, y=0$ [what does this mean?]
How can I proceed to find $y$ in terms of $t$?
I checked out this site but I still don't get how you're supposed to find $y$ in terms of $t$ from it. Some guidance would be appreciated.
 A: I don't know if everything has cleared up for you yet but i wrote the whole thing here if someone still needs it.
Original problem: $$ \dfrac{dy}{dt}=y^2-8y+15 \; , \; y=0 \; t=0 $$
The problem is asking for a particular solution of this ODE at $ y=0 \; t=0$
The dependent and and independent variable is shown like this: $y(t)$, where $y$ depends on $t$ so by plugging in the initial contitions $y=0$ and $t=0\;$ we get $y(0)=0$.
As we get our general solution these initial values should be plugged in and then we can solve our equation for the constant of integration.
By analyzing the specific initial values of our ODE we can find their unique particular solution which by the way can be graphically represented by an integral curve after writing our specific result of the constant of integration back to the general solution.
In some literature $\dfrac{dy}{dt}$ is often marked with $y'(t)$ or $\dot{y}$ .
This is a First Order Inhomogeneous ODE.
This can be solved by separation of variables so integrating both sides with respect to $dy$ and $dt$ respectfully.
$$ \int \dfrac{dy}{y^2-8y+15}= \int dt $$
This LHS integral can be simplified by Partial Fraction Decomposition.
$$\dfrac{1}{y^2-8y+15}= \dfrac{1}{(y-3)(y-5)}= \dfrac{A}{y-3} + \dfrac{B}{y-5}= \dfrac{A(y-5)+B(y-3)}{y^2-8y+15} \Rightarrow$$
where $A, B \in \mathbb{R} $
$$ 1= A(y-5)+B(y-3)= Ay-5A+By-3B$$
$$ y^0: \; 1 = -5A -3B $$
$$ y^1: \; 0 = A+B$$
Solving for $A$ and $B$ we get $A= -\frac{1}{2}$ and $B=\frac{1}{2}$
Substituting back our simplified problem becomes:
$$ \int \left(\dfrac{-\frac{1}{2}}{y-3}+\dfrac{\frac{1}{2}}{y-5}\right)\; dy = \int dt $$
which can be solved by u-substitution or by the $\int \frac{f'}{f}$ rule.
Note: There are 3 integrals actually here, as from the property of linearity we integrate the LHS as two integrals so we'd technically get 3 constants of integration, 2 from $dy$ and 1 from $dt$.
We get:
$$ -\dfrac{1}{2}\ln(|y-3|)+ \dfrac{1}{2}\ln(|y-5|)+C_1 = t+C_2$$
where $C_1, C_2 \in \mathbb{R}$, let $C_2-C_1=C$ which is a real number too.
$$ -\dfrac{1}{2}\ln(|y-3|)+ \dfrac{1}{2}\ln(|y-5|)= t+C$$
$$ -\ln(|y-3|)+ \ln(|y-5|)= 2(t+C)$$
By the quotient rule of logarithms:
$$  \ln\left(\left|\dfrac{y-5}{y-3}\right|\right)= 2(t+C)$$
After exponentiating both sides:
$$ \left|\dfrac{y-5}{y-3}\right| = e^{2(t+C)} $$
Ideally we'd want to solve the equation for $y$ explicitly but this solution is enough in our case.
Let's plug in $y(0)=0$, we can use any equation between $y$ and $t$ after the integrations.
Since we are trying to solve for $C$, let's choose $  \ln\left(\left|\frac{y-5}{y-3}\right|\right)= 2(t+C)$
We get
$$  \ln\left(\left|\dfrac{0-5}{0-3}\right|\right)= 2(0+C) $$
After a bit of cleaning up
$$ C = \dfrac{\ln\left(\dfrac{5}{3}\right)}{2} $$
Which is approximately $$ C \approx 0.25541281188 $$
Let's subtitute $C$ back into our general solution:
$$ \left|\dfrac{y-5}{y-3}\right| = e^{2(t+C)} $$
In an engineering class 2-3 decimal points are usually enough.
So our (approximated) solution is
$$ \boxed{ \left|\dfrac{y-5}{y-3}\right| = e^{2(t+0.256)} } $$
Whose graph looks like this.
For a rather pure/rigourous mathematical setting they'd expect you to give an exact(non-approximated) closed-form solution, which would be:
$$ \boxed{\left|\dfrac{y-5}{y-3}\right| = exp\left({2t+\ln\left(\dfrac{5}{3}\right)}\right)} $$
( exp( for better visibility)
A: ''IF dy/dx is the rate of reaction, then y must be the concentration of product formed'' .
Now, at t=0, y=0 means that when the reaction started( at t=0), the concentration of the products must be zero (y=0)
For rest of the calculation you can refer to @lordcommander 's answer.
That is THE BEST METHOD if you have just got into integrations and differentiations.
