Solving a Differential Equation with Initial Conditions

  • Thread starter snoggerT
  • Start date
In summary, we are looking for a particular solution to the differential equation dy/dx + 2y = 5, with an initial condition of y(0)=0. We can use the method of integrating factors to solve this equation. To start, we need to find a function that satisfies the given equation. From there, we can follow the steps outlined in the conversation to find the unknown function and solve the equation.
  • #1
snoggerT
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Find the particular solution satisfying the initial condition y(0)=0

dy/dx + 2y = 5







The Attempt at a Solution


- I have absolutely no clue where to start. I have a terrible teacher and the book we're using is very poor. If anyone can help me out, that would be great.
 
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  • #3
Can't you give some attempt in solving this? You know the meaning of a derivative, otherwise you would not be working on differential equations, that's a start. Next you know that you should look for a function and not a certain value. So in order to get to this unknown function you have to do something. Can you tell us what?

I can help you through this if you want. Just follow it step by step. First what do you need to do to get to the unknown function?
 

1. What is a particular solution?

A particular solution is a specific set of values for the variables in a general solution that satisfies a given equation or system of equations. It is a unique solution that fulfills all the conditions of the problem.

2. How is a particular solution different from a general solution?

A general solution contains variables and constants, while a particular solution has specific values for those variables and constants. A general solution may have an infinite number of particular solutions.

3. How do you find particular solutions using substitution?

To find a particular solution using substitution, replace the variables in the general solution with the given values and solve for the remaining variable. The resulting solution will be a particular solution that satisfies the given equation.

4. Can a particular solution be a complex number?

Yes, a particular solution can be a complex number. In cases where the given equation involves complex numbers, the particular solution must also be a complex number in order to satisfy the equation.

5. What is the importance of finding particular solutions?

Finding particular solutions is important because it allows us to find specific values that satisfy a given equation or system of equations. These solutions can be used to solve real-world problems and make predictions in various fields such as physics, engineering, and economics.

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