Non homogeneous d.e. solution

  • #1

Homework Statement


hello all,

Suppose y is a solution of the d.e:

y"+p(x)y'+q(x)y= q(x) on the interval (-1,1) with y(0)=1 and y'(0)= 1.

What is y?

Homework Equations



I used the auxiliary equation: m^2+p(x)m+q(x)= q(x)

The Attempt at a Solution


My question is can I do this? I can cancel out the q functions and the right and left and I'm left with m=0 and m=-p(x). I can solve from there and I get a solution of y=1

I am a little uneasy about my solution since I have never seen an auxiliary equation set equal to a function that's not identically zero. Is this legal? Or does my auxiliary equation have to be equal to zero? If that's the case, I don't know how to solve this with the given info since I don't know the particular solution....


Thanks!
 
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Answers and Replies

  • #2
LCKurtz
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Homework Statement


hello all,

Suppose y is a solution of the d.e:

y"+p(x)y'+q(x)y= q(x) on the interval (-1,1) with y(0)=1 and y'(0)= 1.

Is that a typo or is it really the same ##q(x)## on both sides?

What is y?

Homework Equations



I used the auxiliary equation: m^2+p(x)m+q(x)= q(x)

Where did that come from? It makes no sense to me.

The Attempt at a Solution


My question is can I do this? I can cancel out the q functions and the right and left and I'm left with m=0 and m=-p(x). I can solve from there and I get a solution of y=1

Thanks!

Actually, as it's written, ##y \equiv 1## happens to be a solution to the DE, but not the boundary conditions. Back to the drawing board.
 
  • #3
Ray Vickson
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Homework Statement


hello all,

Suppose y is a solution of the d.e:

y"+p(x)y'+q(x)y= q(x) on the interval (-1,1) with y(0)=1 and y'(0)= 1.

What is y?

Homework Equations



I used the auxiliary equation: m^2+p(x)m+q(x)= q(x)

The Attempt at a Solution


My question is can I do this? I can cancel out the q functions and the right and left and I'm left with m=0 and m=-p(x). I can solve from there and I get a solution of y=1

I am a little uneasy about my solution since I have never seen an auxiliary equation set equal to a function that's not identically zero. Is this legal? Or does my auxiliary equation have to be equal to zero? If that's the case, I don't know how to solve this with the given info since I don't know the particular solution....


Thanks!

I suggest you start with the case of constant p and q, and try to solve the DE using both methods you suggested: (i) not cancelling the q on both sides of your m equation; and (ii) cancelling it, as you asked about. One of these methods will yield a horribly incorrect solution, and the best way to understand this is to try it out for yourself.

Also: I bet you have never---in any textbook or paper---seen people write the "auxiliary" equation when p and/or q are functions of x instead of constants. There is a good reason for that!
 
  • #4
vela
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Homework Statement



Suppose y is a solution of the d.e:

y"+p(x)y'+q(x)y= q(x) on the interval (-1,1) with y(0)=1 and y'(0)= 1.

What is y?

Homework Equations



I used the auxiliary equation: m^2+p(x)m+q(x)= q(x)

The Attempt at a Solution


My question is can I do this? I can cancel out the q functions and the right and left and I'm left with m=0 and m=-p(x). I can solve from there and I get a solution of y=1

I am a little uneasy about my solution since I have never seen an auxiliary equation set equal to a function that's not identically zero. Is this legal? Or does my auxiliary equation have to be equal to zero? If that's the case, I don't know how to solve this with the given info since I don't know the particular solution.
You should go back and review the derivation of the method where you used the characteristic polynomial obtained from a differential equation. In particular, pay attention to the assumptions that went into it. Then you can decide whether your approach is valid or not, rather than simply guessing and feeling uneasy.

Oftentimes, it's natural to first concentrate on learning how to use a technique rather than understanding where it came from, but once you've got some familiarity with it, it can be instructive to go back and see the derivation. That way, you'll know when you can use a method and when you can't.
 
  • #5
Is that a typo or is it really the same ##q(x)## on both sides?
yes it's q(x) on both sides. The equation was supposed to be a characteristic equation wich I realize was completely wrong since p(x) and q(x) aren't constant.

Why not try helping me out instead of criticizing me? I'm new to d.e so chill out.
 
  • #6
I suggest you start with the case of constant p and q, and try to solve the DE using both methods you suggested: (i) not cancelling the q on both sides of your m equation; and (ii) cancelling it, as you asked about. One of these methods will yield a horribly incorrect solution, and the best way to understand this is to try it out for yourself.

Also: I bet you have never---in any textbook or paper---seen people write the "auxiliary" equation when p and/or q are functions of x instead of constants. There is a good reason for that!
Thanks for the response! You are completely correct..I forgot that p and q needed to be constant for a characteristic eqn. Im new to second order and haven't even learned to solve a non homogeneous linear yet so the q(x) on the right side is throwing me off....
 
  • #7
LCKurtz
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Why not try helping me out instead of criticizing me? I'm new to d.e so chill out.

Please quote the phrase where I "criticized" you.
 

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