# Victoria's question at Yahoo! Answers regarding a separable first order ODE

#### MarkFL

Staff member
Here is the question:

General solution of dy/dt=k((y)(b-y))?

B is a constant (in this case, initial temperature)
K is a constant of proportionality
I know you have to use separable variables, but i oeep
Getting stuck. Thanks for the help!
Here is a link to the question:

General solution of dy/dt=k((y)(b-y))? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

#### MarkFL

Staff member
Hello Victoria,

We are given to solve:

$$\displaystyle \frac{dy}{dt}=ky(b-y)$$ where $k$ and $b$ are constants.

First, let's separate the variables:

$$\displaystyle \frac{1}{y(b-y)}\,dy=k\,dt$$

Using partial fraction decomposition on the left, and integrating, we obtain:

$$\displaystyle \frac{1}{b}\int\left(\frac{1}{y}-\frac{1}{y-b} \right)\,dy=k\int\,dt$$

$$\displaystyle \ln\left|\frac{y}{y-b} \right|=bkt+C$$

Converting from logarithmic to exponential form, we find:

$$\displaystyle \frac{y}{y-b}=Ce^{bkt}$$

Solving for $y$, we get:

$$\displaystyle y(t)=\frac{b}{1-Ce^{-bkt}}$$ where $$\displaystyle 0<C$$.

To Victoria and any other guests viewing this topic, I invite and encourage you to post of differential equations problems in our Differential Equations forum.

Best Regards,

Mark.

Last edited:

#### chisigma

##### Well-known member
For completeness sake it should be specified that it exists also the solution y=0 that can't be obtained fon any value of the constant C. In general that happens when an ODE has the form $\displaystyle \frac{d y}{d x} = f(y)$ where is $\displaystyle f(0)=0$ ...

Kind regards

$\chi$ $\sigma$

#### MarkFL

I failed to mention that during the process of separation of variables, we in fact did lose the trivial solution $$\displaystyle y \equiv 0$$.