# Indie's question at Yahoo! Answers regarding a separable ODE/partial fractions

#### MarkFL

Staff member
Here is the question:

Partial Decomposition Problem (Calculus)?

I am doing some review problems but I don't understand how to do this problem... any suggestions would be helpful. Thank you in advance. This is for my BC Calculus Class in Nebraska so not too many people are familiar with this math here that I know.

(dy/dx) = .3y - .0001(y^2)
I have posted a link there to this thread so the OP can view my work.

#### MarkFL

Staff member
Hello Indie,

We are given to solve the first order ODE:

$$\displaystyle \frac{dy}{dx}=0.3y-0.0001y^2$$

First, let's multiply through by $10000$ to get rid of the decimals:

$$\displaystyle 10000\frac{dy}{dx}=3000y-y^2$$

Next, let's separate variables:

$$\displaystyle \frac{1}{y(3000-y)}\,dy=\frac{1}{10000}\,dx$$

Now, the expression on the left may be rewritten using partial fraction decomposition. We may assume the decomposition will take the form:

$$\displaystyle \frac{1}{y(3000-y)}=\frac{A}{y}+\frac{B}{3000-y}$$

Multiplying through by $$\displaystyle y(3000-y)$$ we obtain:

$$\displaystyle 1=A(3000-y)+By$$

Since this is true for all $y$, letting $y=0$ we get:

$$\displaystyle 1=3000A\implies A=\frac{1}{3000}$$

And letting $y=3000$, we obtain:

$$\displaystyle 1=3000B\implies B=\frac{1}{3000}$$

Hence:

$$\displaystyle \frac{1}{y(3000-y)}=\frac{1}{3000}\left(\frac{1}{y}+\frac{1}{3000-y} \right)$$

And so the ODE may be written:

$$\displaystyle \left(\frac{1}{y}-\frac{1}{y-3000} \right)\,dy=\frac{3}{10}\,dx$$

Integrating, we find:

$$\displaystyle \int \frac{1}{y}-\frac{1}{y-3000}\,dy=\frac{3}{10}\int\,dx$$

$$\displaystyle \ln\left|\frac{y}{y-3000} \right|=\frac{3}{10}x+C$$

Converting from logarithmic to exponential form (and rewriting the parameter $C$), we have:

$$\displaystyle \frac{y}{y-3000}=Ce^{\frac{3}{10}x}$$

Now we want to solve for $y$. Multiply through by $y-3000$:

$$\displaystyle y=(y-3000)Ce^{\frac{3}{10}x}$$

$$\displaystyle y\left(Ce^{\frac{3}{10}x}-1 \right)=3000Ce^{\frac{3}{10}x}$$

$$\displaystyle y=\frac{3000Ce^{\frac{3}{10}x}}{Ce^{\frac{3}{10}x}-1}$$

Now, dividing each term in the numerator and denominator by $$\displaystyle Ce^{\frac{3}{10}x}$$ and rewriting the parameter again, we obtain:

$$\displaystyle y(x)=\frac{3000}{1+Ce^{-\frac{3}{10}x}}$$