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

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The discussion focuses on solving the first-order ordinary differential equation (ODE) given by (dy/dx) = 0.3y - 0.0001(y^2). The solution involves separating variables and applying partial fraction decomposition to rewrite the equation. The final solution is expressed as y(x) = 3000 / (1 + Ce^(-3/10 x)), where C is a constant determined by initial conditions. This method effectively demonstrates the application of calculus techniques in solving separable ODEs.

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MarkFL
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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.
 
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Hello Indie,

We are given to solve the first order ODE:

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

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

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

Next, let's separate variables:

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

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

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

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

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

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

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

And letting $y=3000$, we obtain:

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

Hence:

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

And so the ODE may be written:

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

Integrating, we find:

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

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

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

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

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

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

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

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

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

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

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