Second order linear differential equation

In summary, the problem asks to solve the second order linear differential equation d2x/dt2 + x = 2 cos(t) with initial conditions x(0) = 0 and dx/dt (0) = 0, and determine the type of motion. The standard method for solving this type of equation is to first solve the associated homogeneous equation and then look for a single solution to the entire equation. In this case, the solution may be of the form y= Ax cos(x)+ Bx sin(x).
  • #1
dvvv
26
0

Homework Statement


Solve the following second order linear differential equation
d2x/dt2 + x = 2 cos(t)
subject to the initial condition x(0) = 0 and dx/dt (0) = 0. What type of motion do you find?



Homework Equations





The Attempt at a Solution


I don't know where to start
 
Physics news on Phys.org
  • #2
If you honestly have no idea how to even start a homework problem, you have a serious problem. I recommend you talk to your teacher about this. Obviously, he/she expects you to know how to do problems like this. If you don't your teacher needs to suggest some review.

The standard method of solving a "non-homogeneous linear differential equation with constant coefficients" is first to solve the associated homogeneous equation:
[tex]\frac{d^2x}{dt^2}+ x= 0[/tex]
Can you do that?

And then look for a single solution to the entire equation. Normally, with "cos(x)" on the left, I would recommend trying something of the form "Acos(x)+ Bsin(x)" but for this equation, as you should be able to see after solving the associated homogeneous equation, that will not work. Instead try something of the form y= Ax cos(x)+ Bx sin(x).
 
  • #3
I think I would substitute x=e^rt. I did that on another question.
I'm probably wrong. It's too late for me to learn all this now. Thanks anyway!
 
  • #4
Is a movement with external forced f(t)=cos(t), without damping, (can be a spring with external force)
 

Related to Second order linear differential equation

1. What is a second order linear differential equation?

A second order linear differential equation is a mathematical expression that describes the relationship between a function and its derivatives. It is in the form of y'' + p(x)y' + q(x)y = g(x), where y is the unknown function, x is the independent variable, and p(x), q(x), and g(x) are known functions.

2. What is the difference between first order and second order differential equations?

The main difference between first and second order differential equations is the highest derivative that appears in the equation. A first order equation has the first derivative as the highest, while a second order equation has the second derivative as the highest. Additionally, second order equations are more complex and require two initial conditions to find a unique solution, while first order equations only require one initial condition.

3. What is the general solution to a second order linear differential equation?

The general solution to a second order linear differential equation is a family of functions that satisfy the equation. It is in the form of y = c1y1(x) + c2y2(x), where y1(x) and y2(x) are linearly independent solutions and c1 and c2 are arbitrary constants.

4. How do you solve a second order linear differential equation with constant coefficients?

To solve a second order linear differential equation with constant coefficients, you can use the method of undetermined coefficients or the method of variation of parameters. The first method involves assuming a particular solution and then finding the general solution using the characteristic equation. The second method involves finding a particular solution by varying the constants in the general solution.

5. What are some real-life applications of second order linear differential equations?

Second order linear differential equations have many practical applications in physics, engineering, and other fields. They can be used to model the motion of a spring, the growth of a population, the vibrations of a guitar string, and many other natural phenomena. They are also used in electrical and mechanical systems to describe the relationship between variables such as current, voltage, and position.

Similar threads

  • Calculus and Beyond Homework Help
Replies
7
Views
435
  • Calculus and Beyond Homework Help
Replies
1
Views
943
  • Calculus and Beyond Homework Help
Replies
2
Views
305
  • Calculus and Beyond Homework Help
Replies
1
Views
386
  • Calculus and Beyond Homework Help
Replies
7
Views
739
  • Calculus and Beyond Homework Help
Replies
2
Views
343
  • Calculus and Beyond Homework Help
Replies
3
Views
146
  • Calculus and Beyond Homework Help
Replies
1
Views
374
  • Calculus and Beyond Homework Help
Replies
14
Views
507
  • Calculus and Beyond Homework Help
Replies
3
Views
626
Back
Top