ODE - Proof of a Particular Solution

In summary, this conversation discusses the use of Green's function as a generalization of variation of parameters in finding a particular solution of a differential equation. It also explains the process of narrowing down the search for a specific solution and solving for the coefficients using algebraic equations.
  • #1
gabriels-horn
92
0

Homework Statement


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Homework Equations


(1) [tex] y\prime\prime+p(x)y\prime+q(x)y=g(x) [/tex]
(2) [tex] y\prime\prime+p(x)y\prime+q(x)y=0 [/tex]

If y1 and y2 are complimentary solutions of (2), then a particular solution of (1) is given by Yp = u1*y1 + u2*y2


The Attempt at a Solution


Anyone have a good starting idea for this one? Perhaps someone with some insight on Green's function?
 
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  • #2
What this problem shows is that Green's function is a generalization of "variation of parameters".

If y1 and y2 are complimentary solutions, then you seek a specific solution of the form y(x)= u(x)y1(x)+ v(x)y2(x). Of course, there are many function, u and v, that will satisfy that. Differentiating, y'= u'y1+ uy1'+ v'y2+ vy2', by the product rule. Because there are many possible solutions, we narrow the search and simplify the problem by requiring that u'y1+ v'y2= 0. Now, we have y'= uy1'+ vy2' so y"= u'y1'+ uy1"+ v'y2'+ vy2". Putting that into the original differential equation, y"+ py'+ qy= u'y1'+ uy1"+ v'y2'+ vy2"+ puy'1+ pvy2'+ quy1+ qvy2= (uy1"+ puy1'+ quy1)+ (vy2"+ pvy2'+ qvy2)+ (u'y1'+ v'uy2')= g(x).

But uy1"+ puy1'+ quy1= u(y1"+ py1'+ qy1)= 0 because y1 satisfies the homogenous equation and vy2"+ pvy2"+ qvy2= v(y1"+ py1'+ qy1)= 0 because y2 satisfies the homogenous equation. That leave u'y1'+ v'y2'= g(x).

Together with u'y1+ v'y2= 0, that give two equations we can solve, algebraically, for u' and v'. Do that, then integrate to get the formula you want.
 
  • #3
HallsofIvy said:
What this problem shows is that Green's function is a generalization of "variation of parameters".

If y1 and y2 are complimentary solutions, then you seek a specific solution of the form y(x)= u(x)y1(x)+ v(x)y2(x). Of course, there are many function, u and v, that will satisfy that. Differentiating, y'= u'y1+ uy1'+ v'y2+ vy2', by the product rule. Because there are many possible solutions, we narrow the search and simplify the problem by requiring that u'y1+ v'y2= 0. Now, we have y'= uy1'+ vy2' so y"= u'y1'+ uy1"+ v'y2'+ vy2". Putting that into the original differential equation, y"+ py'+ qy= u'y1'+ uy1"+ v'y2'+ vy2"+ puy'1+ pvy2'+ quy1+ qvy2= (uy1"+ puy1'+ quy1)+ (vy2"+ pvy2'+ qvy2)+ (u'y1'+ v'uy2')= g(x).

But uy1"+ puy1'+ quy1= u(y1"+ py1'+ qy1)= 0 because y1 satisfies the homogenous equation and vy2"+ pvy2"+ qvy2= v(y1"+ py1'+ qy1)= 0 because y2 satisfies the homogenous equation. That leave u'y1'+ v'y2'= g(x).

Together with u'y1+ v'y2= 0, that give two equations we can solve, algebraically, for u' and v'. Do that, then integrate to get the formula you want.

Great insight. Thank you.
 

FAQ: ODE - Proof of a Particular Solution

1. What is an ODE?

An Ordinary Differential Equation (ODE) is a mathematical equation that describes how one or more unknown functions change over a given independent variable. It is used to model a wide range of natural phenomena in fields such as physics, engineering, and biology.

2. How do you prove a particular solution for an ODE?

To prove a particular solution for an ODE, you must first substitute the proposed solution into the equation and its derivatives. Then, you must show that the solution satisfies the equation for all values of the independent variable. This can be done by using algebraic manipulation and the properties of derivatives.

3. What is the difference between a homogeneous and non-homogeneous ODE?

A homogeneous ODE is one where all terms involve the dependent variable and its derivatives. In other words, there are no external forces or sources affecting the system. On the other hand, a non-homogeneous ODE has additional terms that do not involve the dependent variable or its derivatives, representing external influences on the system.

4. Can an ODE have multiple particular solutions?

Yes, an ODE can have multiple particular solutions. This is because the general solution of an ODE is a combination of a particular solution and the complementary solution, which can have an infinite number of solutions. However, for a given initial condition, there will only be one unique particular solution.

5. What are some real-world applications of ODEs?

ODEs are used to model a wide range of natural phenomena in various fields. Some common applications include describing the motion of objects under the influence of forces, predicting population growth, analyzing electrical circuits, and modeling chemical reactions. They are also used in fields such as economics, epidemiology, and climate science.

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