Solving a Differential Equation: Help Needed!

In summary, the conversation is about solving a simple differential equation involving the variables y, t, m, g, and beta. The approach suggested is to assume a solution for y(t), differentiate it twice, and work out the equation constants. This leads to the equations v = dy/dt and m*dv/dt = -mg - beta*v, which can be integrated to find the solution for y(t). Both the original and alternative forms of the equation are equivalent. One approach to solving this equation involves defining v as dy/dt and using integration to find the solution for y(t).
  • #1
stunner5000pt
1,461
2
can you help me olve this (simple?) differential equation
[tex] m \ddot{y} = -mg - \beta \dot{y} [/tex]
integrate once and i get
[tex] \dot{y} = -gt - \frac{\beta y}{m} + C [/tex]
also can be written as
[tex] \frac{dy}{dt} = -gt - \frac{\beta y}{m} + C [/tex]
both are equivalent

basically trying to get the y on one side adn the t on the other side. HJva tried many ways but can't isolate the two. Any suggestions?
 
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  • #2
Have you tried the approach where you assume a solution for y(t), differentiate it twice and work out the equation constants...? What would be a typical function that could work with this approach...?
 
  • #3
First of all let v = dy/dt, then you have
[itex]m \frac{dv}{dt} = -mg - \beta v[/itex]
From which

[itex] \frac{dv}{g + \frac{\beta}{m} v} = -dt [/itex]

Integrating it now gives

[itex] \frac{m}{\beta} \ln(g + \frac{\beta}{m} v) = -t + C_1 [/itex]

Solving for v gives:

[itex] v = C e^{-\frac{\beta}{m}t} - \frac{mg}{\beta} [/itex]

Where C is some new constant which is I think [itex] C = \frac{m}{\beta}e^{\frac{\beta}{m} C_1} [/itex]. But it doesn't matter though.

Now remember that v = dy/dt. Thus integrating the last equation will give you y(t). (there will be two constants then)
 
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1. How do I know which method to use to solve a differential equation?

There are several methods for solving differential equations, such as separation of variables, substitution, and the method of undetermined coefficients. The best method to use depends on the type of differential equation and its initial conditions. It is important to understand the properties of each method and practice solving different types of equations to determine the most appropriate approach.

2. Can I use a calculator to solve a differential equation?

While calculators can be helpful in solving simple differential equations, it is important to understand the steps involved in solving the equation by hand. Calculators may not always provide accurate or complete solutions, so it is recommended to use them as a tool for checking your work rather than relying on them for the entire solution.

3. What is the purpose of solving a differential equation?

Differential equations are used to model real-world phenomena in many fields of science and engineering. By solving these equations, we can better understand and predict the behavior of systems, such as population growth, chemical reactions, and mechanical systems. They also allow us to make informed decisions and optimize processes in various industries.

4. What are initial conditions and why are they important in solving a differential equation?

Initial conditions refer to the values of the dependent variable and its derivatives at a specific point in time. These conditions are necessary for finding a unique solution to a differential equation. Without them, the solution would have an arbitrary constant, making it impossible to accurately predict the behavior of a system.

5. How can I check if my solution to a differential equation is correct?

One way to check the accuracy of your solution is to plug it back into the original equation and see if it satisfies the equation. You can also use a graphing calculator or software to plot the solution and compare it to the behavior of the system. Additionally, you can verify your solution by taking the derivative and seeing if it matches the original equation.

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