Help finishing a linear differential equation. Mechanics

In summary, we are given an acceleration function d^2x/dt^2 = ge^-kt and we need to find the distance that an object moves in time t. For small t, the result is approximated by x = (gt^2)/2 and for very large t, the speed is approximately constant, known as the terminal speed. To solve for these approximations, we can use the Taylor expansion for the exponential and definite integrals. We can also use the fact that the object starts from rest.
  • #1
navm1
44
0

Homework Statement


Find the distance which an object moves in time t if it starts from rest and has an acceleration d^2x/dt^2 = ge^-kt.
Show that for small t the result is approx "x=(gt^2)/2" and show that for very large t, the speed is approximately constant. the constant is called the terminal speed.

Homework Equations

The Attempt at a Solution


I ended up with v = -(ge^-kt)/k + v_0 and x = (ge^-kt)/k^2 + v_0t + x_0 however I am not sure what to do next. I have tried to solve for k and also set t as zero to get x = g/k^2 but it doesn't seem to be the answer the book is looking for.

Thanks
 
Physics news on Phys.org
  • #2
For small t, look at the Taylor expansion for the exponential.
For large t, take the difference between ##\int_0^{T+\Delta} ge^{-kt}\, dt-\int_0^T ge^{-kt}\, dt ## for large T, or instead of ##T+\Delta,## use ##\infty##.
 
  • #3
##k## is a given parameter. You don't want to solve for it.

You might find it helpful to use definite integrals, e.g.,
$$\int_{v_0}^v \,dv = \int_0^t ge^{-kt}\,dt.$$ The problem statement says the object starts from rest, so use that bit of information too.
 

1. How do I solve a linear differential equation?

To solve a linear differential equation, you first need to separate the variables and then integrate both sides. This will give you the general solution. To find the particular solution, you will need to use initial conditions or boundary conditions.

2. What is the difference between a linear and non-linear differential equation?

A linear differential equation is one in which the dependent variable and its derivatives appear only to the first power and are not multiplied together. A non-linear differential equation is one in which the dependent variable and its derivatives may appear to higher powers and/or be multiplied together.

3. How do I know if a differential equation is separable?

You can determine if a differential equation is separable by checking if the equation can be written in the form of f(x)dx = g(y)dy. If it can, then the equation is separable and can be solved using the method described in the first question.

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

Initial conditions are values given for the dependent variable and/or its derivatives at a specific point. They are important in solving a differential equation because they help determine the particular solution and give a unique solution to the equation.

5. Can I use a calculator to solve a linear differential equation?

While you can use a calculator to check your answer, it is not recommended to solely rely on a calculator to solve a linear differential equation. It is important to understand the concepts and steps involved in solving the equation by hand in order to fully grasp the solution process.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
575
  • Calculus and Beyond Homework Help
Replies
7
Views
280
  • Calculus and Beyond Homework Help
Replies
4
Views
939
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
4K
  • Calculus and Beyond Homework Help
Replies
3
Views
492
  • Calculus and Beyond Homework Help
Replies
1
Views
277
  • Calculus and Beyond Homework Help
Replies
0
Views
162
  • Calculus and Beyond Homework Help
Replies
3
Views
569
  • Calculus and Beyond Homework Help
Replies
5
Views
911
Back
Top