So then is the solution that is finally obtained after adding the "specific" function" a general solution?HallsofIvy said:We first find the general solution to the "associated homogeneous solution" by dropping f(x), which, here, gives y''+ p(x)y'+ y= 0, to get the general solution to that equation, then add a specific function that satisfies the entire equation.
hmm, I am a little confused at this part of Linear ODE. general solution can be found when the equation is y''+ p(x)y'+ y= 0. so I understood this way.HallsofIvy said:by dropping f(x), which, here, gives y''+ p(x)y'+ y= 0, to get the general solution to that equation, then add a specific function that satisfies the entire equation. That is, as I said, the "associated homogeneous equation", NOT the original equation. The fact that f(x) cannot be 0 is irrelevant.
Excellent explanation! I really think that these derivations should be provided in class right after introducing any kind of solutions method so that the student can appreciate the ingenuity behind the generalization immediately.HallsofIvy said:Have you seen the proof that the set of all solutions to an nth order homogeneous linear equation form an n dimensional vector space? That is the fundamental concept behind all such problems!
We can think of a 1 dimensional vector space as a line through the origin of a coordinate system, a 2 dimensional vector space as a plane containing the origin, etc. A "linear manifold" would be a line, plane, etc. that does NOT contain the origin. Such a thing is NOT a vector space because the sum of two "objects" in it is not again in it- addition is not "closed". For example, the graph of y= 3x is a line that contains the origin. Two points on that line are (a, 3a) and (b, 3b). The sum of those two is (a+ b, 3a+ 3b)= (a+ b, 3(a+ b)), again on the line so this set is "closed under addition". Multiplying (a, 3a) by the number p gives (pa, p(3a))= (pa, 3(pa)), again on the line. The points on this line form a "vector space" with that addition and scalar multiplication.
The graph of y= 3x+ 2 is a line that does not contain the origin. The points (a, 3a+ 2) and (b, 3b+ 2) are on that line. Their sum is (a+ b, 3a+ 3b+ 2)= (a+ b. 3(a+ b)+ 2) which is NOT on that line. Similarly, p time (3a, 3a+ 2) gives (p(3a), p(3a)+ 3p) NOT on that line. But if, instead of adding like this, we define sum of (a, 3a+ 1) and (b, 3b+ 1) to be (a+ b, 3(a+ b)+ 1) (geometrically this would be "go from (a, 3a+ 1) and (b, 3b+ 1) to (a, 3a), (b, 3b) on the parallel line through the origin, moving along the mutual perpendicular to those parallel lines, add those, then go back to the original line") we can define an addition of such things (and a similar "scalar multiplication").
We do the same kind of thing with linear differential equations. The set of all solutions to a linear, nth order, homogeneous differential equation form a vector space with the usual sum of functions and multiplication by numbers. If y1 and y2 both satisfy y''+ p(x)y'+ y= 0 so does ay1+ by2 for any numbers a and b: (ay1+ by2)''+ p(x)(ay1+ by2)'+ q(x)(ay1+ by2)= a(y1''+ p(x)y1'+ q(x)y1)+ b(y2''+ p(x)y2'+ q(x)y2)= a(0)+ b(0)= 0.
If, rather, y1 and y2 satisfy y''+ p(x)y'+ q(x)y= f(x), with f(x) NOT 0, ay1+ by2 does not satisfy the equation: (ay1+ by2)''+ p(x)(ay1+ by2)'+ q(x)(ay1+ by2)= a(y1''+ p(x)y1'+ q(x)y1)+ b(y2''+ p(x)y2'+ q(x)y2)= a(f(x)+ b(f(x))= (a+ b)f(x) NOT f(x). Instead suppose y1 and y2 satisfy y''+ p(x)y'+ y= 0 and y3 satisfies y''+ p(x)y'+ y= f(x). Then ay1+ by2+ y3 satisfies y''+ p(x)y'+ q(x)y= f(x) for any a, b: (ay1+ by2+ y3)"+ p(x)(ay1+ by2+ y3)'+ q(x)(y1+ y2+ y3)= (ay1''+ ap(x)y1'+ aq(x)y1)+ (by2''+ bp(x)y2'+ bq(x)y2)+ (y3''+ p(x)y3'+ q(x)y3)= a(0)+ b(0)+ f(x)= f(x).
Further if y1 and y2(x) are solutions to the entire equation, y''+ p(x)y'+ q(x)y= f(x), then y3= y1- y2 satisfies y''+ p(x)y'+ q(x)y= 0 (I will leave that "as an exercise for the reader"). So y1= y3+ y2- a function that satisfies the "associated homogenous equation" plus a function that satisfies the entire equation.
HallsofIvy said:heir sum is (a+ b, 3a+ 3b+ 2)= (a+ b. 3(a+ b)+ 2) which is NOT on that line.
HallsofIvy said:Have you seen the proof that the set of all solutions to an nth order homogeneous linear equation form an n dimensional vector space? That is the fundamental concept behind all such problems!
We can think of a 1 dimensional vector space as a line through the origin of a coordinate system, a 2 dimensional vector space as a plane containing the origin, etc. A "linear manifold" would be a line, plane, etc. that does NOT contain the origin. Such a thing is NOT a vector space because the sum of two "objects" in it is not again in it- addition is not "closed". For example, the graph of y= 3x is a line that contains the origin. Two points on that line are (a, 3a) and (b, 3b). The sum of those two is (a+ b, 3a+ 3b)= (a+ b, 3(a+ b)), again on the line so this set is "closed under addition". Multiplying (a, 3a) by the number p gives (pa, p(3a))= (pa, 3(pa)), again on the line. The points on this line form a "vector space" with that addition and scalar multiplication.
The graph of y= 3x+ 2 is a line that does not contain the origin. The points (a, 3a+ 2) and (b, 3b+ 2) are on that line. Their sum is (a+ b, 3a+ 3b+ 2)= (a+ b. 3(a+ b)+ 2) which is NOT on that line. Similarly, p time (3a, 3a+ 2) gives (p(3a), p(3a)+ 3p) NOT on that line. But if, instead of adding like this, we define sum of (a, 3a+ 1) and (b, 3b+ 1) to be (a+ b, 3(a+ b)+ 1) (geometrically this would be "go from (a, 3a+ 1) and (b, 3b+ 1) to (a, 3a), (b, 3b) on the parallel line through the origin, moving along the mutual perpendicular to those parallel lines, add those, then go back to the original line") we can define an addition of such things (and a similar "scalar multiplication").
We do the same kind of thing with linear differential equations. The set of all solutions to a linear, nth order, homogeneous differential equation form a vector space with the usual sum of functions and multiplication by numbers. If y1 and y2 both satisfy y''+ p(x)y'+ y= 0 so does ay1+ by2 for any numbers a and b: (ay1+ by2)''+ p(x)(ay1+ by2)'+ q(x)(ay1+ by2)= a(y1''+ p(x)y1'+ q(x)y1)+ b(y2''+ p(x)y2'+ q(x)y2)= a(0)+ b(0)= 0.
If, rather, y1 and y2 satisfy y''+ p(x)y'+ q(x)y= f(x), with f(x) NOT 0, ay1+ by2 does not satisfy the equation: (ay1+ by2)''+ p(x)(ay1+ by2)'+ q(x)(ay1+ by2)= a(y1''+ p(x)y1'+ q(x)y1)+ b(y2''+ p(x)y2'+ q(x)y2)= a(f(x)+ b(f(x))= (a+ b)f(x) NOT f(x). Instead suppose y1 and y2 satisfy y''+ p(x)y'+ y= 0 and y3 satisfies y''+ p(x)y'+ y= f(x). Then ay1+ by2+ y3 satisfies y''+ p(x)y'+ q(x)y= f(x) for any a, b: (ay1+ by2+ y3)"+ p(x)(ay1+ by2+ y3)'+ q(x)(y1+ y2+ y3)= (ay1''+ ap(x)y1'+ aq(x)y1)+ (by2''+ bp(x)y2'+ bq(x)y2)+ (y3''+ p(x)y3'+ q(x)y3)= a(0)+ b(0)+ f(x)= f(x).
Further if y1 and y2(x) are solutions to the entire equation, y''+ p(x)y'+ q(x)y= f(x), then y3= y1- y2 satisfies y''+ p(x)y'+ q(x)y= 0 (I will leave that "as an exercise for the reader"). So y1= y3+ y2- a function that satisfies the "associated homogenous equation" plus a function that satisfies the entire equation.
for example,popopopd said:hi, could you explain me why nth order linear ode must have n number of general solutions?is it because we are given with n number of initial conditions to find the general solution, and if so,y_g= ay1(x0)+by2(x0)+cy3(x0) ... nyn(x0)=y0
y_g'=ay1'(x0)+by2'(x0)+cy3(x0)' ... nyn(x0)=y'0
...
and in this condition we can have maximum of n number of solutions because each solution satisfies one specific initial condition)
No. we had the points (a, 3a+ 2) and (b, 3b) (without the added "2"). 3a+ 2+ 3b= 3(a+ b)+ 2.popopopd said:isn't it 3(a+b)+4?
You have asked this question in a separate thread and I answered it there.popopopd said:hi, could you explain me why nth order linear ode must have n number of general solutions?
is it because we are given with n number of initial conditions to find the general solution, and if so,y_g= ay1(x0)+by2(x0)+cy3(x0) ... nyn(x0)=y0
y_g'=ay1'(x0)+by2'(x0)+cy3(x0)' ... nyn(x0)=y'0
...
and in this condition we can have maximum of n number of solutions because each solution satisfies one specific initial condition)
A second order ODE, or ordinary differential equation, is a mathematical equation that involves a function and its derivatives up to the second order. It is commonly used to describe the behavior of physical systems, such as motion, heat transfer, and electrical circuits.
A general solution for a second order ODE is an equation that satisfies the ODE for all possible values of the independent variable. It contains two arbitrary constants, which can be determined by applying initial or boundary conditions to obtain a specific solution.
A second order ODE has a general solution if it is a linear ODE, meaning that the function and its derivatives are only multiplied by constants and added or subtracted. Nonlinear ODEs, on the other hand, often do not have a general solution and require numerical methods to find an approximate solution.
The existence of a general solution for a second order ODE is crucial in understanding and predicting the behavior of physical systems. It allows us to find an exact solution for a given set of conditions and to study the long-term behavior of the system. Additionally, many physical laws and equations can be represented as second order ODEs, making the concept of a general solution essential in various areas of science and engineering.
No, a second order ODE can only have one general solution. However, it can have multiple specific solutions that satisfy the ODE for different sets of initial or boundary conditions. These specific solutions can be obtained by substituting different values for the arbitrary constants in the general solution.