Solving Systems of Linear Differential Equations

[shelovesmath]

Hi all.

Woohoo, I'm in diff eq now. Gosh, I've been on this board since Calculus I think. Sorry I usually only come around when I have a question. :\

Sooooo, Thursday my professor did an example of solving a system of linear differential equations with using the linear operator D.

There was no explanation involved, just "Here's how you solve this one."

So, now I'm doing my homework, and I need to clarify, and this might seem like a silly question, but bear with me.

Is the whole idea just to get the system into a form that is solvable only by linear methods?

For example, if you have a system with 2 equations, in 2 unknowns, say x and y, and you eliminate y, you get it into a form where you will solve for x, then rinse, lather, repeat, go back eliminate and solve for y.

If after eliminating x or y, it can be any order, right? And, if it's first order, it will be a linear or Bernoulli form, if it's second order, it will be undetermined coefficients or variation of parameters form, etc.

It seems to me that since it's a linear system, the solution should ONLY be solvable by linear means. No exact, no separable, no homogeneous method. Just linear methods.

Is this correct?

 

[HallsofIvy]

Hi all.

Woohoo, I'm in diff eq now. Gosh, I've been on this board since Calculus I think. Sorry I usually only come around when I have a question. :\

Sooooo, Thursday my professor did an example of solving a system of linear differential equations with using the linear operator D.

There was no explanation involved, just "Here's how you solve this one."

So, now I'm doing my homework, and I need to clarify, and this might seem like a silly question, but bear with me.

Is the whole idea just to get the system into a form that is solvable only by linear methods?

For example, if you have a system with 2 equations, in 2 unknowns, say x and y, and you eliminate y, you get it into a form where you will solve for x, then rinse, lather, repeat, go back eliminate and solve for y.

If after eliminating x or y, it can be any order, right? And, if it's first order, it will be a linear or Bernoulli form, if it's second order, it will be undetermined coefficients or variation of parameters form, etc.

It seems to me that since it's a linear system, the solution should ONLY be solvable by linear means. No exact, no separable, no homogeneous method. Just linear methods.

Is this correct?

I have no idea what you mean by that last paragraph. If you have 2 linear differential equations in the two unknown functions, x and y, you can eliminate one, say y, to get a single second order linear differential equation in the other function, x.

"Exact", "separable", and "homogenous" apply to first order equations, not higher order. However, you need to be aware that, unfortunately, the term "homogeneous" is now used in a different way: a linear equation of higher order is said to be "homogeneous" if every term involves the dependent function, y, or one of its derivatives. That has nothing to do with the concept of "homogeneous" for first order equations.

 

[shelovesmath]

I have no idea what you mean by that last paragraph. If you have 2 linear differential equations in the two unknown functions, x and y, you can eliminate one, say y, to get a single second order linear differential equation in the other function, x.

"Exact", "separable", and "homogenous" apply to first order equations, not higher order. However, you need to be aware that, unfortunately, the term "homogeneous" is now used in a different way: a linear equation of higher order is said to be "homogeneous" if every term involves the dependent function, y, or one of its derivatives. That has nothing to do with the concept of "homogeneous" for first order equations.

I'm sorry, it was a poorly worded question.
If I have a differential equation specifically of the form:

a1(t)x' +a2(t)y' + a3(t)x + a4(t)y = f(t)
b1(t)x' +b2(t)y' + b3(t)x + b4(t)y = g(t)

which my book calls a special case of a linear system of differential equations, when specifically solving it by using the differential operator x' = Dx and y' = Dy, the goal is to use the elimination method to isolate x, find a general solution, then use elimination method again to find y, find another general solution, and in the end, make sure the number of scalars are equal to the order of the differential equations you got after using elimination.

What I'm curious about are those differential equations you get after doing the elimination. If you end up with a first order differential equation, will it always be linear or bernoulli? Is it even possible to end up with a first order differential equation after using the elimination method?