Why 2nd order differential equation has two solutions

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Discussion Overview

The discussion revolves around the nature of second-order differential equations, specifically focusing on why they typically have two independent solutions in the case of homogeneous equations with zero source terms. Participants explore the implications of boundary conditions and the dimensionality of the solution space.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants assert that a second-order linear differential equation of the form y'' + f(x)y' + g(x)y = 0 has two independent solutions if f and g are continuous on the interval considered.
  • One participant presents a counterexample involving a nonlinear equation, stating that it can have four independent solutions, thus challenging the generalization about second-order equations.
  • Another participant emphasizes that every linear homogeneous second-order differential equation has two independent solutions because the solution set forms a vector space of dimension n, where n is the order of the equation.
  • Participants discuss the relationship between independent boundary conditions and the solutions of the differential equation, questioning whether specifying certain initial conditions makes others independent.
  • There is a mention of the need for careful consideration when discussing independence in the context of nonlinear equations.

Areas of Agreement / Disagreement

Participants express differing views on the nature of solutions to second-order differential equations, particularly regarding linear versus nonlinear cases. There is no consensus on the implications of boundary conditions or the generalization about the number of independent solutions.

Contextual Notes

Limitations include the dependence on the definitions of linearity and independence, as well as the unresolved nature of the implications of nonlinear equations in this context.

goodphy
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Hello.

I forgot the reason why 2nd order differential equation has two independent solutions. (Here, source term is zero) Why 3 or 4 independent solutions are not possible?

Please give me clear answer.
 
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goodphy said:
I forgot the reason why 2nd order differential equation has two independent solutions. (Here, source term is zero) Why 3 or 4 independent solutions are not possible?
How many independent boundary conditions can there be a 2nd order differential equation? If someone specifies x(0) and x'(0), is x''(0) independent from those?
 
goodphy said:
Hello.

I forgot the reason why 2nd order differential equation has two independent solutions. (Here, source term is zero) Why 3 or 4 independent solutions are not possible?

Please give me clear answer.

You need to be a little careful. The statement as you've stated it is not correct. (This is a common error).

Consider the equation \left(y''+y\right) \left(y''-y \right)=0

You can check that this equation is 2nd order by expanding the left side:
\left(y''\right)^2-y^2=0

The highest derivative of y is 2nd order. However this equation has 4 independent solutions:
y = c_1 e^x + c_2 e^{-x} + c_3 cos x + c_4 sin x

You should note that this equation is also nonlinear.

A correct statement is that a second order linear differential equation of the form y'' + f(x)y' + g(x) y =0 has two independent solutions on an interval I if f and g are both continuous on the interval I.

This theorem can be generalized to higher dimensions.

A proof of the above statement is given in Tenenbaum and Pollard Lesson 65. In my opinion this is a great reference for differential equations.

You can probably also find the proof online using some google-fu if you search for general solution of 2nd order (or n-th order) differential equations.
 
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jbriggs444 said:
How many independent boundary conditions can there be a 2nd order differential equation? If someone specifies x(0) and x'(0), is x''(0) independent from those?

I see.

2nd order linear differential equation without source (homogeneous linear differential equation) can be always written as y'' + P(x)y' + Q(x)y = 0 where ' means differentiation. Full solution is obtained by solving this equation with initial conditions (equation describes evolution of the system from initial conditions). If we know, y(0) and y'(0) then the equation tells y''(0). These initial conditions gives next time values y(t1), y'(t1) and substituting these into the equation tells y''(t1). Next time values are obtained in similar process so full solution is obtained.

As coefficients of independent solutions are rooms for initial conditions and we need only two conditions to fully describe system evolution thus 2nd order homogeneous differential equation only requires two independent solutions as constituents of general solutions.

Thanks to remind me of this!
 
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Every linear homogeneous second order differential equation has two independent solution because the set of all solutions to an nth order linear homogenous equation is a vector space of dimension n. If you want to include non-linear equations, you will have to specify what you mean by "independent".
 

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