Finding Homogeneous Solutions for Differential Equations: Does Your Guess Work?

[jimmianlin]

Homework Statement

Given a differential equation.
ie. y'' +y' + 1 =0 (THIS IS NOT THE PROBLEM THAT I AM SOLVING)

Homework Equations

No equations

The Attempt at a Solution

The equation above is not what I'm working with, but an example of a differential equation problem that I was working on. Now, I am trying to find a homogeneous solution by guessing a value for y. However, when I make a guess for y and plug it in, the equation gives me 0 = 1. Does this mean the homogenous solution does not exist or did I not make the correct guess?

 

[rock.freak667]

For a differential equation of the form ay''+by'+cy=0 where a,b,c are constants; all solutions are of the form y=e^rx

 

[Dick]

For a differential equation of the form ay''+by'+cy=0 where a,b,c are constants; all solutions are of the form y=e^rx

That's a little over simplistic. E.g. what about y''=0? jimmianlin, what problem are you trying to solve and are you really supposed to solve it by guessing? Guessing y=constant probably won't get you very far.

 

[Pengwuino]

The equation above is not what I'm working with, but an example of a differential equation problem that I was working on. Now, I am trying to find a homogeneous solution by guessing a value for y. However, when I make a guess for y and plug it in, the equation gives me 0 = 1. Does this mean the homogenous solution does not exist or did I not make the correct guess?

The homogeneous part of your solution solves the homogeneous part of your problem. Your homogeneous solution, y[x], will solve the problem of y''[x] + y'[x] = 0. Remember, the "1" is the inhomogeneous part so it isn't part of what you're trying to solve for your homogeneous solution.

 

[rock.freak667]

That's a little over simplistic. E.g. what about y''=0?

Doesn't it work for that as well? The characteristic equation would be r²=0, giving roots 0,0. So y=(Ax+B)e^0x→y=Ax+B

same way you'd get from y''=0 to y'=A to y=Ax+B

 

[Dick]

Doesn't it work for that as well? The characteristic equation would be r²=0, giving roots 0,0. So y=(Ax+B)e^0x→y=Ax+B

same way you'd get from y''=0 to y'=A to y=Ax+B

I'm just saying Ax+B does not have the form e^rx. That's all.

 

[HallsofIvy]

You got in ahead of me, Dick! But it's a very useful oversimplification! Looking for solutions of the form $e^rx$ leads to the characteristic equation that then leads to other solutions such as polynomials, sine and cosine, and combinations of those with exponentials. rock.freak667 oversimplified his language a little.