Solve second order differential equation

In summary, the given function \phi(t) = Asin(wt) + Bcos(wt) is a solution of the second order differential equation \ddot{\phi} = -w^{2}\phi. This is considered the general solution since it involves two arbitrary constants. Direct substitution is required to show that the second derivative of the given function is equal to the negative square of w times the given function. This involves differentiating the function twice and reducing it to an identity that is valid for all t. The differentiation process uses the fact that wt is not treated as a normal variable, but as a constant.
  • #1
Oblio
398
0
Solve by direct substitution taht the function [tex]\phi[/tex](t) = Asin(wt) + Bcos(wt) is a solution of the second order differential equation [tex]\ddot{\phi}[/tex] = -w[tex]^{2}[/tex][tex]\phi[/tex]. ( Since this solution involves two arbitrary constants - the coeffecients of the sine and consine functions - it is in fact the general solution).

I'm not sure what is meant by direct substitution..
Can someone help me get started?
Thanks alot!
 
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  • #2
What does the word "substitution" normally mean?
 
  • #3
Is direct substitution different then normal?
I tried subbing in Asin(wt) + Bcos(wt) into the second order differential equation but that got me nowhere...
 
  • #4
Well, you are asked to show that the second derivative of the given function equals the negative square of w times the given function.

Have you tried to show that?
 
  • #5
That's why I'm confsued.
Am i substituting or differentiating?
 
  • #6
Eeh?

You are to show that by SUBSTITUTING into the diff.eq, which requires you to perform differentiation twice, the differential equation reduces to to an IDENTITY, valid for all t, i.e, that the given function is, indeed, a SOLUTION to the differential equation.
 
  • #7
So I am differentiating

[tex]\ddot{\phi}[/tex]= -w[tex]^{2}[/tex](Asin(wt) + Bcos(wt))
 
  • #8
Quite so!

Does that equal [itex]-w^{2}\phi[/itex]??
 
  • #9
I hope so, it probably will turn out not to be so when I do it wrong :)
 
  • #10
(long derivative..)
 
  • #11
wt wouldn't be constants... not treated as normal variables?
 

FAQ: Solve second order differential equation

What is a second order differential equation?

A second order differential equation is a mathematical equation that involves a function and its first and second derivatives. It is typically written in the form of y'' + p(x)y' + q(x)y = r(x), where y is the unknown function and p(x), q(x), and r(x) are given functions. It is commonly used in physics and engineering to model systems that involve acceleration and velocity.

How do you solve a second order differential equation?

To solve a second order differential equation, you first need to identify the type of equation you are dealing with. There are three main types: homogeneous, non-homogeneous, and exact. Next, you can use various methods such as the method of undetermined coefficients, variation of parameters, or the Laplace transform to find the solution. It is important to follow the steps carefully and check your work for accuracy.

What are the initial conditions in a second order differential equation?

The initial conditions in a second order differential equation refer to the values of the unknown function and its derivatives at a specific point. Typically, two initial conditions are needed to find a unique solution to the equation. These conditions can be given in the form of y(a) = b and y'(a) = c, where a is the point and b and c are constants.

Can a second order differential equation have multiple solutions?

Yes, a second order differential equation can have multiple solutions. This depends on the type of equation and the initial conditions given. It is also possible for an equation to have no solution or an infinite number of solutions. It is important to carefully consider the initial conditions and the methods used to solve the equation to find the correct solution.

How is a second order differential equation used in real life applications?

A second order differential equation is used in many real life applications, particularly in fields such as physics, engineering, and economics. It is used to model systems that involve acceleration and velocity, such as the motion of objects under the influence of gravity. It is also used to describe the behavior of electric circuits, mass-spring systems, and other physical systems. In economics, it is used to model growth and change in various economic variables.

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