# Simple Differential Equation

• RentonT
In summary: OK. So what would be on the left-hand side of the equal sign?x=\frac{A}{y}+\frac{B}{1000-y}I understand you multiply the left side by one of the denominators on the right side and then plug in the x value that would have zeroed the denominator. I just don't know what goes on the left side. Is it the \frac{dy}{dt}?Yes, the left side is the \frac{dy}{dt} .

## Homework Statement

"Solve this differential equation algebraically, subject to the initial condition that $$y=10$$ at $$t=0$$

## Homework Equations

$$\frac{dy}{dt} = 2y*\frac{1000-y}{1000}$$

## The Attempt at a Solution

I first reduced the right side to $$\frac{-y^2}{500} + 2y$$
After that I separated the variables, but I don't know how to integrate that function.
Can anyone point me in the right step?
$$\int \frac {dy}{\frac{-y^2}{500}+2y}$$

Last edited:
Have you done any partial fractions recently? You can write the integrand as A/(y)+B/(1000-y). You just need to find A and B.

Dick said:
Have you done any partial fractions recently? You can write the integrand as A/(y)+B/(1000-y). You just need to find A and B.
I am in fact working on the section over partial fractions. Thank you very much Dick. I see how partial fractions plays into this. Might I ask how you got $$\frac{A}{y} + \frac{B}{1000-y}$$

I'm having trouble seeing the partial fractions. Most of them I have worked with I use the shortcut to Heaviside's Method. Many have been in the form of $$\int \frac {Linear}{Quadratic}$$

Last edited:
RentonT said:
I am in fact working on the section over partial fractions. Thank you very much Dick. I see how partial fractions plays into this. Might I ask how you got $$\frac{A}{y} + \frac{B}{1000-y}$$

y and (1000-y) were the factors of your original expression before you multiplied it out. You didn't really want to do that.

Dick said:
y and (1000-y) were the factors of your original expression before you multiplied it out. You didn't really want to do that.

OK. So what would be on the left-hand side of the equal sign?
$$x=\frac{A}{y}+\frac{B}{1000-y}$$

I understand you multiply the left side by one of the denominators on the right side and then plug in the x value that would have zeroed the denominator. I just don't know what goes on the left side. Is it the $$\frac{dy}{dt}$$?

Nevermind. My poor algebra skills were my downfall this time. I figured out that it's $$\int \frac{-500}{y(y-1000)}$$

## 1. What is a simple differential equation?

A simple differential equation is an equation that describes the relationship between a function and its derivatives. It contains one independent variable and one dependent variable, and the goal is to find the function that satisfies the equation.

## 2. What is the difference between an ordinary and a partial differential equation?

Ordinary differential equations involve only one independent variable, while partial differential equations involve multiple independent variables. Additionally, the derivatives in ordinary differential equations are with respect to the single independent variable, while in partial differential equations, they can be with respect to any of the independent variables.

## 3. How is a simple differential equation solved?

A simple differential equation is typically solved by finding a general solution using integration and then applying initial or boundary conditions to find a particular solution that satisfies the given conditions.

## 4. What are some real-life applications of simple differential equations?

Simple differential equations are used to model a wide range of physical phenomena, such as population growth, chemical reactions, and electrical circuits. They are also used in economics, biology, and engineering to make predictions and solve problems.

## 5. Are there any limitations to using simple differential equations?

Yes, there are some limitations to using simple differential equations. They may not be able to accurately model complex systems with multiple independent variables and nonlinear relationships. In these cases, more advanced techniques, such as numerical methods, may be necessary.