Characteristic equation with x^2 coefficient

  • #1
knockout_artist
70
2

Homework Statement



x2 d2y/dx2 + 3x dy/dx + 5y = g(x)

Homework Equations


How do we find Characteristic equation for it.

The Attempt at a Solution



x2λ2 + 3xλ + 5 = 0
λ1 = 1/2 [-x2 + √ (x4 + 20 ) ]
λ2 = 1/2[ -x2 - √(x4 + 20) ]

I used 1/3 -/+ a √(a2 + 4b)
where
a = x2
b = 5
 
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  • #2
knockout_artist said:

Homework Statement



x2 d2y/dx2 + 3x dy/dx + 5y = g(x)

Homework Equations


How do we find Characteristic equation for it.
If memory serves, the characteristic equation applies to the related nonhomogeneous DE -- ##x^2 y'' + 3x y' + 5y = 0##.
This is an example of an 2nd order Euler equation. One technique is to assume a solution of the form ##y = x^n##, and substitute it into the DE. For your nonhomogeneous equation, it depends on what g(x) is.
See https://www.math24.net/second-order-euler-equation/
knockout_artist said:

The Attempt at a Solution



x2λ2 + 3xλ + 5 = 0
λ1 = 1/2 [-x2 + √ (x4 + 20 ) ]
I have no idea what you're doing here.
knockout_artist said:
λ2 = 1/2[ -x2 - √(x4 + 20) ]

I used 1/3 -/+ a √(a2 + 4b)
where
a = x2
b = 5
 
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  • #3
knockout_artist said:

Homework Statement



x2 d2y/dx2 + 3x dy/dx + 5y = g(x)

Homework Equations


How do we find Characteristic equation for it.

The Attempt at a Solution



x2λ2 + 3xλ + 5 = 0
λ1 = 1/2 [-x2 + √ (x4 + 20 ) ]
λ2 = 1/2[ -x2 - √(x4 + 20) ]

I used 1/3 -/+ a √(a2 + 4b)
where
a = x2
b = 5

Forget characteristic equations: they do not apply in this problem. They are for constant coefficient linear DEs, but your DE has coefficients that are functions of ##x##.
 
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  • #4
Ray Vickson said:
Forget characteristic equations: they do not apply in this problem.
The old saying applies here: "If the only tool you have is a hammer, everything looks like a nail."
As Ray said (and I forgot), characteristic equations are applicable only to constant coefficient linear differential equations, and specifically to homogeneous DEs of that type.
 
  • #5
If you need a quick reference, the method is called cauchy-euler (I think this is the name). There is also an equivalent form using a log, worth it to know both versions. t I believe you can also use Frobineous Method, due to the analytical point. Its been years since I solved a differential equation.
 
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